Companion to the University-Style Practice Set大学水平练习题配套解答
Sections 1 to 6: trig integrals (sin/cos/sec/tan powers), trig substitution (all three cases), partial fractions (linear, repeated, irreducible quadratic), rationalising substitutions, and integration strategyCALC II第 1 至 6 节：三角积分（sin/cos/sec/tan 的幂次）、三角换元法（三种情形）、部分分数分解（线性因子、重复因子、不可约二次因子）、有理化换元，以及积分策略CALC II
Evaluate: (a) $\int \sin^4 x\,dx$; (b) $\int \sin^3 x \cos^4 x\,dx$; (c) $\int \tan^4 x\,dx$.计算：(a) $\int \sin^4 x\,dx$；(b) $\int \sin^3 x \cos^4 x\,dx$；(c) $\int \tan^4 x\,dx$。
Both exponents are even, so use the half-angle identity $\sin^2 x = \tfrac{1}{2}(1 - \cos 2x)$: (M1)两个指数均为偶数，故使用半角恒等式 $\sin^2 x = \tfrac{1}{2}(1 - \cos 2x)$：(M1)
$$ \sin^4 x = \left(\frac{1-\cos 2x}{2}\right)^2 = \frac{1 - 2\cos 2x + \cos^2 2x}{4}. $$Apply the half-angle identity a second time to $\cos^2 2x = \tfrac{1}{2}(1 + \cos 4x)$: (M1)对 $\cos^2 2x = \tfrac{1}{2}(1 + \cos 4x)$ 再次应用半角恒等式：(M1)
$$ \sin^4 x = \frac{1}{4}\left(1 - 2\cos 2x + \frac{1+\cos 4x}{2}\right) = \frac{3}{8} - \frac{\cos 2x}{2} + \frac{\cos 4x}{8}. $$Integrating term by term:逐项积分：
$$ \int \sin^4 x\,dx = \frac{3x}{8} - \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + C. $$Verification by differentiation: $\frac{d}{dx}\!\left(\frac{3x}{8} - \frac{\sin 2x}{4} + \frac{\sin 4x}{32}\right) = \frac{3}{8} - \frac{\cos 2x}{2} + \frac{\cos 4x}{8} = \sin^4 x$. (A1)微分验证：$\frac{d}{dx}\!\left(\frac{3x}{8} - \frac{\sin 2x}{4} + \frac{\sin 4x}{32}\right) = \frac{3}{8} - \frac{\cos 2x}{2} + \frac{\cos 4x}{8} = \sin^4 x$。(A1)
The power of sine is odd. Reserve one factor of $\sin x$ for $du = -\sin x\,dx$ and write $\sin^2 x = 1 - \cos^2 x$: (M1)正弦的次幂为奇数。保留一个 $\sin x$ 用于 $du = -\sin x\,dx$，并将 $\sin^2 x = 1 - \cos^2 x$ 代入：(M1)
$$ \int \sin^3 x \cos^4 x\,dx = \int (1 - \cos^2 x)\cos^4 x \cdot \sin x\,dx. $$Let $u = \cos x$, $du = -\sin x\,dx$: (M1)令 $u = \cos x$，$du = -\sin x\,dx$：(M1)
$$ -\int (1 - u^2)u^4\,du = -\int (u^4 - u^6)\,du = -\frac{u^5}{5} + \frac{u^7}{7} + C = -\frac{\cos^5 x}{5} + \frac{\cos^7 x}{7} + C. $$Verification: $\frac{d}{dx}\!\left(-\frac{\cos^5 x}{5} + \frac{\cos^7 x}{7}\right) = \cos^4 x\sin x - \cos^6 x\sin x = \sin x\cos^4 x(1 - \cos^2 x) = \sin^3 x\cos^4 x$. (A1)验证：$\frac{d}{dx}\!\left(-\frac{\cos^5 x}{5} + \frac{\cos^7 x}{7}\right) = \cos^4 x\sin x - \cos^6 x\sin x = \sin x\cos^4 x(1 - \cos^2 x) = \sin^3 x\cos^4 x$。(A1)
Write $\tan^4 x = \tan^2 x(\sec^2 x - 1) = \tan^2 x\sec^2 x - \tan^2 x$. For the second piece apply the identity again: $\tan^2 x = \sec^2 x - 1$. (M1)将 $\tan^4 x = \tan^2 x(\sec^2 x - 1) = \tan^2 x\sec^2 x - \tan^2 x$。对第二项再次应用恒等式：$\tan^2 x = \sec^2 x - 1$。(M1)
$$ \int \tan^4 x\,dx = \int \tan^2 x\sec^2 x\,dx - \int (\sec^2 x - 1)\,dx. $$The first integral uses $u = \tan x$, $du = \sec^2 x\,dx$, giving $\frac{\tan^3 x}{3}$. The second integrates directly:第一个积分令 $u = \tan x$，$du = \sec^2 x\,dx$，得 $\frac{\tan^3 x}{3}$。第二个直接积分：
$$ \int \tan^4 x\,dx = \frac{\tan^3 x}{3} - \tan x + x + C. $$Verification: $\frac{d}{dx}\!\left(\frac{\tan^3 x}{3} - \tan x + x\right) = \tan^2 x\sec^2 x - \sec^2 x + 1 = \sec^2 x(\tan^2 x - 1) + 1 = \sec^2 x(\tan^2 x - 1) + 1$. Using $\sec^2 x = 1 + \tan^2 x$: $(1+\tan^2 x)(\tan^2 x - 1) + 1 = \tan^4 x - 1 + 1 = \tan^4 x$. (A1)验证：$\frac{d}{dx}\!\left(\frac{\tan^3 x}{3} - \tan x + x\right) = \tan^2 x\sec^2 x - \sec^2 x + 1 = \sec^2 x(\tan^2 x - 1) + 1$。利用 $\sec^2 x = 1 + \tan^2 x$：$(1+\tan^2 x)(\tan^2 x - 1) + 1 = \tan^4 x - 1 + 1 = \tan^4 x$。(A1)
Evaluate: (a) $\int \tan^3 x\sec^4 x\,dx$; (b) $\int \sec^6 x\,dx$.计算：(a) $\int \tan^3 x\sec^4 x\,dx$；(b) $\int \sec^6 x\,dx$。
The secant power is even, so reserve one $\sec^2 x$ for $du = \sec^2 x\,dx$ with $u = \tan x$, and convert the remaining $\sec^2 x = 1 + \tan^2 x$: (M1)正割次幂为偶数，保留一个 $\sec^2 x$ 用于 $du = \sec^2 x\,dx$，令 $u = \tan x$，并将剩余部分 $\sec^2 x = 1 + \tan^2 x$ 代入：(M1)
$$ \int \tan^3 x \sec^4 x\,dx = \int \tan^3 x(1 + \tan^2 x)\sec^2 x\,dx. $$Let $u = \tan x$: (M1)令 $u = \tan x$：(M1)
$$ \int u^3(1 + u^2)\,du = \int (u^3 + u^5)\,du = \frac{u^4}{4} + \frac{u^6}{6} + C = \frac{\tan^4 x}{4} + \frac{\tan^6 x}{6} + C. $$Verification: $\frac{d}{dx}\!\left(\frac{\tan^4 x}{4} + \frac{\tan^6 x}{6}\right) = \tan^3 x\sec^2 x + \tan^5 x\sec^2 x = \sec^2 x\tan^3 x(1 + \tan^2 x) = \tan^3 x\sec^4 x$. (A1)验证：$\frac{d}{dx}\!\left(\frac{\tan^4 x}{4} + \frac{\tan^6 x}{6}\right) = \tan^3 x\sec^2 x + \tan^5 x\sec^2 x = \sec^2 x\tan^3 x(1 + \tan^2 x) = \tan^3 x\sec^4 x$。(A1)
Reserve $\sec^2 x\,dx = du$ with $u = \tan x$ and write $\sec^4 x = (1 + \tan^2 x)^2 = 1 + 2\tan^2 x + \tan^4 x$: (M1)令 $u = \tan x$，保留 $\sec^2 x\,dx = du$，并将 $\sec^4 x = (1 + \tan^2 x)^2 = 1 + 2\tan^2 x + \tan^4 x$ 代入：(M1)
$$ \int \sec^6 x\,dx = \int \sec^4 x \cdot \sec^2 x\,dx = \int (1 + 2\tan^2 x + \tan^4 x)\sec^2 x\,dx. $$With $u = \tan x$: (M1)令 $u = \tan x$：(M1)
$$ \int (1 + 2u^2 + u^4)\,du = u + \frac{2u^3}{3} + \frac{u^5}{5} + C = \tan x + \frac{2\tan^3 x}{3} + \frac{\tan^5 x}{5} + C. $$Verification: $\frac{d}{dx}\!\left(\tan x + \frac{2\tan^3 x}{3} + \frac{\tan^5 x}{5}\right) = \sec^2 x + 2\tan^2 x\sec^2 x + \tan^4 x\sec^2 x = \sec^2 x(1 + 2\tan^2 x + \tan^4 x) = \sec^2 x(1+\tan^2 x)^2 = \sec^6 x$. (A1)验证：$\frac{d}{dx}\!\left(\tan x + \frac{2\tan^3 x}{3} + \frac{\tan^5 x}{5}\right) = \sec^2 x + 2\tan^2 x\sec^2 x + \tan^4 x\sec^2 x = \sec^2 x(1 + 2\tan^2 x + \tan^4 x) = \sec^2 x(1+\tan^2 x)^2 = \sec^6 x$。(A1)
Evaluate: (a) $\int \frac{x^2}{\sqrt{9-x^2}}\,dx$; (b) $\int \frac{dx}{(4+x^2)^2}$.计算：(a) $\int \frac{x^2}{\sqrt{9-x^2}}\,dx$；(b) $\int \frac{dx}{(4+x^2)^2}$。
The form $\sqrt{9 - x^2} = \sqrt{3^2 - x^2}$ calls for $x = 3\sin\theta$, $dx = 3\cos\theta\,d\theta$, $\sqrt{9 - x^2} = 3\cos\theta$ on $-\tfrac{\pi}{2}\le\theta\le\tfrac{\pi}{2}$. (M1)$\sqrt{9 - x^2} = \sqrt{3^2 - x^2}$ 的形式要求令 $x = 3\sin\theta$，$dx = 3\cos\theta\,d\theta$，$\sqrt{9 - x^2} = 3\cos\theta$，其中 $-\tfrac{\pi}{2}\le\theta\le\tfrac{\pi}{2}$。(M1)
$$ \int \frac{9\sin^2\theta}{3\cos\theta}\cdot 3\cos\theta\,d\theta = 9\int \sin^2\theta\,d\theta = 9\int \frac{1-\cos 2\theta}{2}\,d\theta = \frac{9\theta}{2} - \frac{9\sin 2\theta}{4} + C. $$(M1) Now $\sin 2\theta = 2\sin\theta\cos\theta$. From the reference triangle (opposite $x$, hypotenuse $3$, adjacent $\sqrt{9-x^2}$): $\sin\theta = x/3$, $\cos\theta = \sqrt{9-x^2}/3$, and $\theta = \arcsin(x/3)$. (A1)(M1) 现有 $\sin 2\theta = 2\sin\theta\cos\theta$。由参考三角形（对边 $x$，斜边 $3$，邻边 $\sqrt{9-x^2}$）得：$\sin\theta = x/3$，$\cos\theta = \sqrt{9-x^2}/3$，$\theta = \arcsin(x/3)$。(A1)
$$ = \frac{9}{2}\arcsin\frac{x}{3} - \frac{9}{4}\cdot 2\cdot\frac{x}{3}\cdot\frac{\sqrt{9-x^2}}{3} + C = \frac{9}{2}\arcsin\frac{x}{3} - \frac{x\sqrt{9-x^2}}{2} + C. $$(A1) Verification: $\tfrac{d}{dx}\arcsin\tfrac{x}{3}=\tfrac{1}{\sqrt{1-x^2/9}}\cdot\tfrac{1}{3}=\tfrac{1}{\sqrt{9-x^2}}$, so $\tfrac{d}{dx}\bigl(\tfrac{9}{2}\arcsin\tfrac{x}{3}\bigr)=\tfrac{9}{2\sqrt{9-x^2}}$. Also $\tfrac{d}{dx}\bigl(-\tfrac{x\sqrt{9-x^2}}{2}\bigr)=-\tfrac{\sqrt{9-x^2}}{2}+\tfrac{x^2}{2\sqrt{9-x^2}}=\tfrac{2x^2-9}{2\sqrt{9-x^2}}$. Sum: $\tfrac{9}{2\sqrt{9-x^2}}+\tfrac{2x^2-9}{2\sqrt{9-x^2}}=\tfrac{2x^2}{2\sqrt{9-x^2}}=\tfrac{x^2}{\sqrt{9-x^2}}$. Confirmed.(A1) 验证：$\tfrac{d}{dx}\arcsin\tfrac{x}{3}=\tfrac{1}{\sqrt{1-x^2/9}}\cdot\tfrac{1}{3}=\tfrac{1}{\sqrt{9-x^2}}$，故 $\tfrac{d}{dx}\bigl(\tfrac{9}{2}\arcsin\tfrac{x}{3}\bigr)=\tfrac{9}{2\sqrt{9-x^2}}$。又 $\tfrac{d}{dx}\bigl(-\tfrac{x\sqrt{9-x^2}}{2}\bigr)=-\tfrac{\sqrt{9-x^2}}{2}+\tfrac{x^2}{2\sqrt{9-x^2}}=\tfrac{2x^2-9}{2\sqrt{9-x^2}}$。两项相加：$\tfrac{9}{2\sqrt{9-x^2}}+\tfrac{2x^2-9}{2\sqrt{9-x^2}}=\tfrac{x^2}{\sqrt{9-x^2}}$。验证正确。
The form $(4 + x^2)^2$ calls for $x = 2\tan\theta$, $dx = 2\sec^2\theta\,d\theta$, $4 + x^2 = 4\sec^2\theta$. (M1)$(4 + x^2)^2$ 的形式要求令 $x = 2\tan\theta$，$dx = 2\sec^2\theta\,d\theta$，$4 + x^2 = 4\sec^2\theta$。(M1)
$$ \int \frac{2\sec^2\theta\,d\theta}{16\sec^4\theta} = \frac{1}{8}\int \cos^2\theta\,d\theta = \frac{1}{8}\int \frac{1+\cos 2\theta}{2}\,d\theta = \frac{\theta}{16} + \frac{\sin 2\theta}{32} + C. $$(M1) From the reference triangle (opposite $x$, adjacent $2$, hypotenuse $\sqrt{4+x^2}$): $\theta = \arctan(x/2)$, $\sin\theta = x/\sqrt{4+x^2}$, $\cos\theta = 2/\sqrt{4+x^2}$, and $\sin 2\theta = 2\sin\theta\cos\theta = 4x/(4+x^2)$. (A1)(M1) 由参考三角形（对边 $x$，邻边 $2$，斜边 $\sqrt{4+x^2}$）得：$\theta = \arctan(x/2)$，$\sin\theta = x/\sqrt{4+x^2}$，$\cos\theta = 2/\sqrt{4+x^2}$，$\sin 2\theta = 2\sin\theta\cos\theta = 4x/(4+x^2)$。(A1)
$$ = \frac{1}{16}\arctan\frac{x}{2} + \frac{4x}{32(4+x^2)} + C = \frac{1}{16}\arctan\frac{x}{2} + \frac{x}{8(4+x^2)} + C. $$(A1) Verification at $x=0$: antiderivative is $0$, and the integrand is $1/16 > 0$, consistent with $d/dx(\frac{1}{16}\arctan 0 + 0) = \frac{1}{16}\cdot\frac{1}{2} + \frac{1}{8}\cdot\frac{4-0}{16} = \frac{1}{32} + \frac{1}{32} = \frac{1}{16} = \frac{1}{(4+0)^2}$. Correct.(A1) 在 $x=0$ 处验证：原函数为 $0$，被积函数为 $1/16 > 0$，与 $d/dx(\frac{1}{16}\arctan 0 + 0) = \frac{1}{32} + \frac{1}{32} = \frac{1}{16} = \frac{1}{(4+0)^2}$ 一致。正确。
Evaluate: (a) $\int \frac{3x+1}{x^2-x-6}\,dx$; (b) $\int \frac{5x^2-3x-2}{x(x-1)(x+2)}\,dx$.计算：(a) $\int \frac{3x+1}{x^2-x-6}\,dx$；(b) $\int \frac{5x^2-3x-2}{x(x-1)(x+2)}\,dx$。
Factor: $x^2 - x - 6 = (x-3)(x+2)$. Write (M1)因式分解：$x^2 - x - 6 = (x-3)(x+2)$。写出 (M1)
$$ \frac{3x+1}{(x-3)(x+2)} = \frac{A}{x-3} + \frac{B}{x+2}. $$Clear denominators: $3x + 1 = A(x+2) + B(x-3)$. Set $x=3$: $10 = 5A$, so $A=2$. Set $x=-2$: $-5 = -5B$, so $B=1$.去分母：$3x + 1 = A(x+2) + B(x-3)$。令 $x=3$：$10 = 5A$，故 $A=2$。令 $x=-2$：$-5 = -5B$，故 $B=1$。
Verification by recombining: $\dfrac{2}{x-3}+\dfrac{1}{x+2} = \dfrac{2(x+2)+(x-3)}{(x-3)(x+2)} = \dfrac{3x+1}{x^2-x-6}$. Confirmed. (A1)通分验证：$\dfrac{2}{x-3}+\dfrac{1}{x+2} = \dfrac{2(x+2)+(x-3)}{(x-3)(x+2)} = \dfrac{3x+1}{x^2-x-6}$。验证正确。(A1)
$$ \int \frac{3x+1}{x^2-x-6}\,dx = 2\ln|x-3| + \ln|x+2| + C. \quad\text{(A1)} $$Write (M1)写出 (M1)
$$ \frac{5x^2-3x-2}{x(x-1)(x+2)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+2}. $$Clear denominators: $5x^2-3x-2 = A(x-1)(x+2) + Bx(x+2) + Cx(x-1)$.去分母：$5x^2-3x-2 = A(x-1)(x+2) + Bx(x+2) + Cx(x-1)$。
Cover-up: Set $x=0$: $-2 = A(-1)(2)$, so $A=1$. (A1) Set $x=1$: $5-3-2=0 = B(1)(3)$, so $B=0$. Set $x=-2$: $20+6-2=24 = C(-2)(-3)=6C$, so $C=4$.覆盖法：令 $x=0$：$-2 = A(-1)(2)$，故 $A=1$。(A1) 令 $x=1$：$5-3-2=0 = B(1)(3)$，故 $B=0$。令 $x=-2$：$20+6-2=24 = C(-2)(-3)=6C$，故 $C=4$。
Verification: $\dfrac{1}{x}+\dfrac{4}{x+2} = \dfrac{(x-1)(x+2)+4x(x-1)}{x(x-1)(x+2)} = \dfrac{(x-1)(5x+2)}{x(x-1)(x+2)} = \dfrac{5x^2-3x-2}{x(x-1)(x+2)}$. Confirmed. (A1)验证：$\dfrac{1}{x}+\dfrac{4}{x+2} = \dfrac{(x-1)(x+2)+4x(x-1)}{x(x-1)(x+2)} = \dfrac{5x^2-3x-2}{x(x-1)(x+2)}$。验证正确。(A1)
$$ \int \frac{5x^2-3x-2}{x(x-1)(x+2)}\,dx = \ln|x| + 4\ln|x+2| + C. $$(a) Show $\sin^4 x = \tfrac{3}{8} - \tfrac{1}{2}\cos 2x + \tfrac{1}{8}\cos 4x$ using only $\sin^2 x = \tfrac{1}{2}(1-\cos 2x)$; (b) hence find $\int_0^{\pi/2}\sin^4 x\,dx$; (c) deduce $\int_0^{\pi/2}\sin^4 x\,dx = \int_0^{\pi/2}\cos^4 x\,dx$ without recomputing the right side.(a) 仅用 $\sin^2 x = \tfrac{1}{2}(1-\cos 2x)$ 证明 $\sin^4 x = \tfrac{3}{8} - \tfrac{1}{2}\cos 2x + \tfrac{1}{8}\cos 4x$；(b) 由此求 $\int_0^{\pi/2}\sin^4 x\,dx$；(c) 无须重新计算右边，推断 $\int_0^{\pi/2}\sin^4 x\,dx = \int_0^{\pi/2}\cos^4 x\,dx$。
Starting from $\sin^2 x = \tfrac{1}{2}(1 - \cos 2x)$, square both sides: (M1)从 $\sin^2 x = \tfrac{1}{2}(1 - \cos 2x)$ 出发，对两边平方：(M1)
$$ \sin^4 x = \left(\frac{1-\cos 2x}{2}\right)^2 = \frac{1 - 2\cos 2x + \cos^2 2x}{4}. $$Apply the same identity a second time to the $\cos^2 2x$ term: replace $x$ by $2x$ to give $\cos^2 2x = \tfrac{1}{2}(1 + \cos 4x)$. (M1)对 $\cos^2 2x$ 项再次应用同一恒等式：将 $x$ 替换为 $2x$ 得 $\cos^2 2x = \tfrac{1}{2}(1 + \cos 4x)$。(M1)
$$ \sin^4 x = \frac{1}{4}\left(1 - 2\cos 2x + \frac{1 + \cos 4x}{2}\right) = \frac{1}{4}\cdot\frac{2 - 4\cos 2x + 1 + \cos 4x}{2} = \frac{3 - 4\cos 2x + \cos 4x}{8}. $$(A1) Writing out the three terms:(A1) 写出三项：
$$ \sin^4 x = \frac{3}{8} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x. \quad\text{(A1)} $$No other identities were assumed; both applications used only $\cos^2\alpha = \tfrac{1}{2}(1+\cos 2\alpha)$ with $\alpha = x$ then $\alpha = 2x$.未假设其他恒等式；两次均仅使用 $\cos^2\alpha = \tfrac{1}{2}(1+\cos 2\alpha)$，分别取 $\alpha = x$ 和 $\alpha = 2x$。
The antiderivative is (M1)原函数为 (M1)
$$ F(x) = \frac{3x}{8} - \frac{\sin 2x}{4} + \frac{\sin 4x}{32}. $$Evaluate:代入积分限：
$$ F\!\left(\frac{\pi}{2}\right) = \frac{3\pi}{16} - \frac{\sin\pi}{4} + \frac{\sin 2\pi}{32} = \frac{3\pi}{16} - 0 + 0 = \frac{3\pi}{16}. $$ $$ F(0) = 0. $$Hence $\displaystyle\int_0^{\pi/2}\sin^4 x\,dx = \frac{3\pi}{16}$. (A1)故 $\displaystyle\int_0^{\pi/2}\sin^4 x\,dx = \frac{3\pi}{16}$。(A1)
Use the substitution $x = \tfrac{\pi}{2} - t$, $dx = -dt$ in the right-hand integral: (R1)在右边积分中令 $x = \tfrac{\pi}{2} - t$，$dx = -dt$：(R1)
$$ \int_0^{\pi/2}\cos^4 x\,dx = \int_{\pi/2}^{0}\cos^4\!\left(\frac{\pi}{2}-t\right)(-dt) = \int_0^{\pi/2}\sin^4 t\,dt. $$Since $\cos\!\left(\frac{\pi}{2}-t\right) = \sin t$, the two integrals are identical. (A1) Alternatively, the expansion $\cos^4 x = \tfrac{3}{8} + \tfrac{1}{2}\cos 2x + \tfrac{1}{8}\cos 4x$ differs from the $\sin^4 x$ expansion only in the sign of the $\cos 2x$ term, and $\int_0^{\pi/2}\cos 2x\,dx = \tfrac{1}{2}[\sin 2x]_0^{\pi/2} = 0$, so the definite integrals coincide regardless.因 $\cos\!\left(\frac{\pi}{2}-t\right) = \sin t$，两个积分完全相同。(A1) 另一方面，展开式 $\cos^4 x = \tfrac{3}{8} + \tfrac{1}{2}\cos 2x + \tfrac{1}{8}\cos 4x$ 与 $\sin^4 x$ 的展开式仅 $\cos 2x$ 项的符号不同，而 $\int_0^{\pi/2}\cos 2x\,dx = \tfrac{1}{2}[\sin 2x]_0^{\pi/2} = 0$，故两个定积分相等。
For $R(x) = \dfrac{4x^3+3x^2+8x+3}{(x+1)^2(x^2+3)}$: (a) write the general PF form and justify each term; (b) determine all four coefficients; (c) verify by recombining.对于 $R(x) = \dfrac{4x^3+3x^2+8x+3}{(x+1)^2(x^2+3)}$：(a) 写出一般部分分数形式并说明每项的依据；(b) 求出全部四个系数；(c) 通分还原验证。
$(x+1)^2$ is a repeated linear factor: it contributes one term per power, i.e. $\tfrac{A}{x+1}$ and $\tfrac{B}{(x+1)^2}$. (A1)$(x+1)^2$ 是重复线性因子：每个次幂各贡献一项，即 $\tfrac{A}{x+1}$ 和 $\tfrac{B}{(x+1)^2}$。(A1)
$x^2+3$ is irreducible over $\mathbb{R}$ (discriminant $-12<0$): it contributes a linear numerator $Cx+D$. (A1)$x^2+3$ 在 $\mathbb{R}$ 上不可约（判别式 $-12<0$）：贡献线性分子 $Cx+D$。(A1)
$$ R(x) = \frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{Cx+D}{x^2+3}. $$Multiply both sides by $(x+1)^2(x^2+3)$: (M1)两边同乘 $(x+1)^2(x^2+3)$：(M1)
$$ 4x^3+3x^2+8x+3 = A(x+1)(x^2+3)+B(x^2+3)+(Cx+D)(x+1)^2. $$Expand each piece and collect by degree: (M1)展开各项并按次数归类：(M1)
$$ A(x+1)(x^2+3) = Ax^3+Ax^2+3Ax+3A, $$ $$ B(x^2+3) = Bx^2+3B, $$ $$ (Cx+D)(x+1)^2 = Cx^3+(2C+D)x^2+(C+2D)x+D. $$Matching powers: $[x^3]:\ A+C=4$; $[x^2]:\ A+B+2C+D=3$; $[x^1]:\ 3A+C+2D=8$; $[x^0]:\ 3A+3B+D=3$. (M1)比较各次系数：$[x^3]:\ A+C=4$；$[x^2]:\ A+B+2C+D=3$；$[x^1]:\ 3A+C+2D=8$；$[x^0]:\ 3A+3B+D=3$。(M1)
From $[x^3]$ and $[x^1]$: $C=4-A$ and $3A+(4-A)+2D=8\Rightarrow A+D=2\Rightarrow D=2-A$. (A1)由 $[x^3]$ 和 $[x^1]$：$C=4-A$，且 $3A+(4-A)+2D=8\Rightarrow A+D=2\Rightarrow D=2-A$。(A1)
Substituting into $[x^2]$: $A+B+2(4-A)+(2-A)=3\Rightarrow B-2A=-7\Rightarrow B=2A-7$.代入 $[x^2]$：$A+B+2(4-A)+(2-A)=3\Rightarrow B-2A=-7\Rightarrow B=2A-7$。
Substituting into $[x^0]$: $3A+3(2A-7)+(2-A)=3\Rightarrow 8A=22\Rightarrow A=\tfrac{11}{4}$.代入 $[x^0]$：$3A+3(2A-7)+(2-A)=3\Rightarrow 8A=22\Rightarrow A=\tfrac{11}{4}$。
Hence $C=\tfrac{5}{4}$, $D=-\tfrac{3}{4}$, $B=-\tfrac{3}{2}$. (A1)故 $C=\tfrac{5}{4}$，$D=-\tfrac{3}{4}$，$B=-\tfrac{3}{2}$。(A1)
Multiply out the numerator contributions over the common denominator: (M1)在公分母下展开各分子贡献：(M1)
$$ \frac{11}{4}(x+1)(x^2+3) = \frac{11x^3+11x^2+33x+33}{4}, $$ $$ -\frac{3}{2}(x^2+3) = \frac{-6x^2-18}{4}, $$ $$ \frac{1}{4}(5x-3)(x^2+2x+1) = \frac{5x^3+7x^2-x-3}{4}. $$Summing by degree: (A1)按次数求和：(A1)
$$ [x^3]:\ \frac{11+5}{4}=4;\quad [x^2]:\ \frac{11-6+7}{4}=3;\quad [x^1]:\ \frac{33-1}{4}=8;\quad [x^0]:\ \frac{33-18-3}{4}=3. $$Recombined numerator: $4x^3+3x^2+8x+3$. Confirmed. (A1)还原后的分子：$4x^3+3x^2+8x+3$。验证正确。(A1)
Evaluate $\int \frac{dx}{x^2\sqrt{x^2-16}}$ for $x>4$.计算 $\int \frac{dx}{x^2\sqrt{x^2-16}}$，其中 $x>4$。
The form $\sqrt{x^2 - 16} = \sqrt{x^2 - 4^2}$ calls for $x = 4\sec\theta$, with $a = 4$. (M1)$\sqrt{x^2 - 16} = \sqrt{x^2 - 4^2}$ 的形式要求令 $x = 4\sec\theta$，其中 $a = 4$。(M1)
Then $dx = 4\sec\theta\tan\theta\,d\theta$ and, for $0 \le \theta < \tfrac{\pi}{2}$ (i.e. $x > 4$), $\sqrt{x^2 - 16} = \sqrt{16\sec^2\theta - 16} = 4|\tan\theta| = 4\tan\theta$. (M1)则 $dx = 4\sec\theta\tan\theta\,d\theta$，且在 $0 \le \theta < \tfrac{\pi}{2}$（即 $x > 4$）时，$\sqrt{x^2 - 16} = \sqrt{16\sec^2\theta - 16} = 4|\tan\theta| = 4\tan\theta$。(M1)
$$ \int \frac{4\sec\theta\tan\theta\,d\theta}{16\sec^2\theta \cdot 4\tan\theta} = \frac{1}{16}\int \frac{d\theta}{\sec\theta} = \frac{1}{16}\int \cos\theta\,d\theta. \quad\text{(A1)} $$$\dfrac{1}{16}\int \cos\theta\,d\theta = \dfrac{\sin\theta}{16} + C.$ (M1·A1)$\dfrac{1}{16}\int \cos\theta\,d\theta = \dfrac{\sin\theta}{16} + C$。(M1·A1)
Draw the reference triangle for $x = 4\sec\theta$: (M1)画出 $x = 4\sec\theta$ 的参考三角形：(M1)
Read off: $\sin\theta = \dfrac{\sqrt{x^2-16}}{x}$. (M1)从三角形读出：$\sin\theta = \dfrac{\sqrt{x^2-16}}{x}$。(M1)
$$ \int \frac{dx}{x^2\sqrt{x^2-16}} = \frac{\sin\theta}{16} + C = \frac{\sqrt{x^2-16}}{16x} + C. $$Verification by differentiation: $\dfrac{d}{dx}\!\left(\dfrac{\sqrt{x^2-16}}{16x}\right) = \dfrac{1}{16}\cdot\dfrac{x\cdot\frac{x}{\sqrt{x^2-16}} - \sqrt{x^2-16}}{x^2} = \dfrac{1}{16x^2}\cdot\dfrac{x^2-(x^2-16)}{\sqrt{x^2-16}} = \dfrac{16}{16x^2\sqrt{x^2-16}} = \dfrac{1}{x^2\sqrt{x^2-16}}$. (A1)微分验证：$\dfrac{d}{dx}\!\left(\dfrac{\sqrt{x^2-16}}{16x}\right) = \dfrac{1}{16}\cdot\dfrac{x\cdot\frac{x}{\sqrt{x^2-16}} - \sqrt{x^2-16}}{x^2} = \dfrac{1}{16x^2}\cdot\dfrac{16}{\sqrt{x^2-16}} = \dfrac{1}{x^2\sqrt{x^2-16}}$。(A1)
Evaluate $\int_0^1 \frac{x^2}{\sqrt{x^2-2x+5}}\,dx$.计算 $\int_0^1 \frac{x^2}{\sqrt{x^2-2x+5}}\,dx$。
$x^2 - 2x + 5 = (x-1)^2 + 4$. (M1) This is of the form $u^2 + 4$ with $u = x-1$, calling for the tangent substitution $u = 2\tan\theta$, i.e. $x - 1 = 2\tan\theta$, $dx = 2\sec^2\theta\,d\theta$, $\sqrt{(x-1)^2+4} = 2\sec\theta$. (A1)$x^2 - 2x + 5 = (x-1)^2 + 4$。(M1) 令 $u = x-1$，这是 $u^2 + 4$ 的形式，需使用正切换元 $u = 2\tan\theta$，即 $x - 1 = 2\tan\theta$，$dx = 2\sec^2\theta\,d\theta$，$\sqrt{(x-1)^2+4} = 2\sec\theta$。(A1)
New limits: when $x=0$, $\tan\theta = -\tfrac{1}{2}$, so $\theta = \arctan(-\tfrac{1}{2}) = -\arctan\tfrac{1}{2}$. When $x=1$, $\tan\theta = 0$, so $\theta = 0$. (A1)新积分限：当 $x=0$ 时，$\tan\theta = -\tfrac{1}{2}$，故 $\theta = -\arctan\tfrac{1}{2}$；当 $x=1$ 时，$\tan\theta = 0$，故 $\theta = 0$。(A1)
With $x = 1 + 2\tan\theta$, the numerator $x^2 = (1+2\tan\theta)^2 = 1 + 4\tan\theta + 4\tan^2\theta$. (M1)令 $x = 1 + 2\tan\theta$，分子 $x^2 = (1+2\tan\theta)^2 = 1 + 4\tan\theta + 4\tan^2\theta$。(M1)
$$ \int \frac{(1+4\tan\theta+4\tan^2\theta)\cdot 2\sec^2\theta\,d\theta}{2\sec\theta} = \int (1+4\tan\theta+4\tan^2\theta)\sec\theta\,d\theta. $$Split into three integrals: (M1)拆分为三个积分：(M1)
$$ I_1 = \int \sec\theta\,d\theta = \ln|\sec\theta+\tan\theta|, $$ $$ I_2 = 4\int \tan\theta\sec\theta\,d\theta = 4\sec\theta, $$ $$ I_3 = 4\int \tan^2\theta\sec\theta\,d\theta. $$For $I_3$, use $\tan^2\theta = \sec^2\theta - 1$: $I_3 = 4\int(\sec^3\theta - \sec\theta)\,d\theta$. Recall $\int\sec^3\theta\,d\theta = \tfrac{1}{2}(\sec\theta\tan\theta + \ln|\sec\theta+\tan\theta|)$: (M1)对 $I_3$，用 $\tan^2\theta = \sec^2\theta - 1$：$I_3 = 4\int(\sec^3\theta - \sec\theta)\,d\theta$。利用 $\int\sec^3\theta\,d\theta = \tfrac{1}{2}(\sec\theta\tan\theta + \ln|\sec\theta+\tan\theta|)$：(M1)
$$ I_3 = 4\left[\frac{\sec\theta\tan\theta + \ln|\sec\theta+\tan\theta|}{2} - \ln|\sec\theta+\tan\theta|\right] = 2\sec\theta\tan\theta - 2\ln|\sec\theta+\tan\theta|. $$Total antiderivative: $F(\theta) = \ln|\sec\theta+\tan\theta| + 4\sec\theta + 2\sec\theta\tan\theta - 2\ln|\sec\theta+\tan\theta|$总原函数：$F(\theta) = \ln|\sec\theta+\tan\theta| + 4\sec\theta + 2\sec\theta\tan\theta - 2\ln|\sec\theta+\tan\theta|$
$$ = 4\sec\theta + 2\sec\theta\tan\theta - \ln|\sec\theta+\tan\theta|. $$(A1) Now evaluate between $\theta = -\arctan\tfrac{1}{2}$ and $\theta = 0$. At $\theta=0$: $\sec 0=1$, $\tan 0=0$: $F(0) = 4 + 0 - \ln 1 = 4$. (A1)(A1) 在 $\theta = -\arctan\tfrac{1}{2}$ 至 $\theta = 0$ 间求值。当 $\theta=0$ 时：$\sec 0=1$，$\tan 0=0$，故 $F(0) = 4 + 0 - \ln 1 = 4$。(A1)
At $\theta=-\arctan\tfrac{1}{2}$: $\tan\theta = -\tfrac{1}{2}$, $\sec\theta = \sqrt{1+\tfrac{1}{4}} = \tfrac{\sqrt{5}}{2}$:当 $\theta=-\arctan\tfrac{1}{2}$ 时：$\tan\theta = -\tfrac{1}{2}$，$\sec\theta = \sqrt{1+\tfrac{1}{4}} = \tfrac{\sqrt{5}}{2}$：
$$ F(-\arctan\tfrac{1}{2}) = 4\cdot\frac{\sqrt{5}}{2} + 2\cdot\frac{\sqrt{5}}{2}\cdot\left(-\frac{1}{2}\right) - \ln\left|\frac{\sqrt{5}}{2} - \frac{1}{2}\right| = 2\sqrt{5} - \frac{\sqrt{5}}{2} - \ln\frac{\sqrt{5}-1}{2}. $$The definite integral equals $F(0) - F(-\arctan\tfrac{1}{2})$: (A1)定积分等于 $F(0) - F(-\arctan\tfrac{1}{2})$：(A1)
$$ = 4 - 2\sqrt{5} + \frac{\sqrt{5}}{2} + \ln\frac{\sqrt{5}-1}{2} = 4 - \frac{3\sqrt{5}}{2} + \ln\frac{\sqrt{5}-1}{2}. $$This is positive: numerically $4 - \tfrac{3\times2.236}{2} + \ln(0.618) \approx 4 - 3.354 - 0.481 \approx 0.165 > 0$. (A1)此值为正：数值上 $4 - \tfrac{3\times2.236}{2} + \ln(0.618) \approx 4 - 3.354 - 0.481 \approx 0.165 > 0$。(A1)
Find the area between $y=\dfrac{6x^2+4}{(x^2+1)(x^2+4)}$ and $y=0$ for $0\le x\le 2$.求曲线 $y=\dfrac{6x^2+4}{(x^2+1)(x^2+4)}$ 与 $y=0$ 在 $0\le x\le 2$ 之间所围区域的面积。
Each factor $x^2+1$ and $x^2+4$ is irreducible over the reals (negative discriminants). Since the numerator degree (2) is less than the denominator degree (4), the fraction is proper. Write: (M1)$x^2+1$ 和 $x^2+4$ 在实数域上均不可约（判别式为负）。由于分子次数（2）小于分母次数（4），该分式为真分式。写出：(M1)
$$ \frac{6x^2+4}{(x^2+1)(x^2+4)} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+4}. $$Because the integrand is an even function of $x$ (the numerator and denominator both contain only even powers), the partial-fraction decomposition must also be even. Therefore $A = C = 0$, and we need only find $B$ and $D$: (M1)由于被积函数是 $x$ 的偶函数（分子和分母均只含偶数次幂），部分分数分解也必须为偶函数，故 $A = C = 0$，只需求 $B$ 和 $D$：(M1)
$$ \frac{6x^2+4}{(x^2+1)(x^2+4)} = \frac{B}{x^2+1} + \frac{D}{x^2+4}. $$Clear denominators: $6x^2+4 = B(x^2+4) + D(x^2+1)$. Matching coefficients: $[x^2]: B+D=6$; $[x^0]: 4B+D=4$. Subtracting: $3B=-2$, so $B=-\tfrac{2}{3}$, $D=\tfrac{20}{3}$. (A1)去分母：$6x^2+4 = B(x^2+4) + D(x^2+1)$。比较系数：$[x^2]: B+D=6$；$[x^0]: 4B+D=4$。相减得 $3B=-2$，故 $B=-\tfrac{2}{3}$，$D=\tfrac{20}{3}$。(A1)
Verification: $-\tfrac{2}{3}(x^2+4)+\tfrac{20}{3}(x^2+1) = \frac{-2x^2-8+20x^2+20}{3} = \frac{18x^2+12}{3} = 6x^2+4$. Confirmed. (A1)验证：$-\tfrac{2}{3}(x^2+4)+\tfrac{20}{3}(x^2+1) = \frac{18x^2+12}{3} = 6x^2+4$。验证正确。(A1)
(M1) Use $\int\frac{dx}{x^2+a^2} = \frac{1}{a}\arctan\frac{x}{a}+C$: (M1)(M1) 利用 $\int\frac{dx}{x^2+a^2} = \frac{1}{a}\arctan\frac{x}{a}+C$：(M1)
$$ = \left[-\frac{2}{3}\arctan x + \frac{20}{3}\cdot\frac{1}{2}\arctan\frac{x}{2}\right]_0^2 = \left[-\frac{2}{3}\arctan x + \frac{10}{3}\arctan\frac{x}{2}\right]_0^2. $$At $x=2$: $-\tfrac{2}{3}\arctan 2 + \tfrac{10}{3}\arctan 1 = -\tfrac{2}{3}\arctan 2 + \tfrac{10}{3}\cdot\tfrac{\pi}{4} = -\tfrac{2}{3}\arctan 2 + \tfrac{5\pi}{6}$. (A1)当 $x=2$ 时：$-\tfrac{2}{3}\arctan 2 + \tfrac{10}{3}\arctan 1 = -\tfrac{2}{3}\arctan 2 + \tfrac{5\pi}{6}$。(A1)
At $x=0$: $0 + 0 = 0$.当 $x=0$ 时：$0 + 0 = 0$。
Area $= \dfrac{5\pi}{6} - \dfrac{2}{3}\arctan 2$. (A1)面积 $= \dfrac{5\pi}{6} - \dfrac{2}{3}\arctan 2$。(A1)
Area $= \dfrac{5\pi}{6} - \dfrac{2}{3}\arctan 2$. (A1)面积 $= \dfrac{5\pi}{6} - \dfrac{2}{3}\arctan 2$。(A1)
Numerically: $\arctan 2 \approx 1.107$, so $\dfrac{2}{3}(1.107)\approx 0.738$ and $\dfrac{5\pi}{6}\approx 2.618$. Area $\approx 2.618 - 0.738 = 1.880 > 0$. The integrand is positive for $x\in[0,2]$, so the area is indeed positive. (A1)数值上：$\arctan 2 \approx 1.107$，故 $\dfrac{2}{3}(1.107)\approx 0.738$，$\dfrac{5\pi}{6}\approx 2.618$，面积 $\approx 2.618 - 0.738 = 1.880 > 0$。被积函数在 $x\in[0,2]$ 上为正，故面积确实为正。(A1)
Evaluate $\int \frac{1}{1+\sqrt{x+1}}\,dx$ using $u=\sqrt{x+1}$.使用 $u=\sqrt{x+1}$ 计算 $\int \frac{1}{1+\sqrt{x+1}}\,dx$。
Let $u = \sqrt{x+1}$, so $u^2 = x+1$ and $x = u^2 - 1$, $dx = 2u\,du$. (M1) Substitute: (M1)令 $u = \sqrt{x+1}$，则 $u^2 = x+1$，$x = u^2 - 1$，$dx = 2u\,du$。(M1) 代入：(M1)
$$ \int \frac{1}{1+u}\cdot 2u\,du = \int \frac{2u}{1+u}\,du. $$The integrand is a rational function of $u$. (A1)被积函数是 $u$ 的有理函数。(A1)
The degree of the numerator equals the degree of the denominator, so perform polynomial long division: (M1)分子次数等于分母次数，故进行多项式长除法：(M1)
$$ \frac{2u}{1+u} = \frac{2u}{u+1} = 2 - \frac{2}{u+1}. $$Check: $2(u+1) - 2 = 2u$. Confirmed. The proper-fraction part is $-\tfrac{2}{u+1}$, which is already in partial-fraction form (one linear factor in the denominator). (A1)验证：$2(u+1) - 2 = 2u$。正确。真分式部分为 $-\tfrac{2}{u+1}$，已是部分分数形式（分母为一个线性因子）。(A1)
$$ \int \left(2 - \frac{2}{u+1}\right)du = 2u - 2\ln|u+1| + C. $$(A1) Verification: $\frac{d}{du}(2u - 2\ln|u+1|) = 2 - \frac{2}{u+1} = \frac{2u}{u+1}$. Confirmed.(A1) 验证：$\frac{d}{du}(2u - 2\ln|u+1|) = 2 - \frac{2}{u+1} = \frac{2u}{u+1}$。正确。
Since $x \ge -1$ we have $u = \sqrt{x+1} \ge 0$, so $u+1 \ge 1 > 0$ and the absolute value is unnecessary: (M1)由于 $x \ge -1$，有 $u = \sqrt{x+1} \ge 0$，故 $u+1 \ge 1 > 0$，绝对值符号可省略：(M1)
$$ \int \frac{dx}{1+\sqrt{x+1}} = 2\sqrt{x+1} - 2\ln(1+\sqrt{x+1}) + C. $$Verification: let $F(x) = 2\sqrt{x+1} - 2\ln(1+\sqrt{x+1})$. Then验证：令 $F(x) = 2\sqrt{x+1} - 2\ln(1+\sqrt{x+1})$，则
$$ F'(x) = \frac{1}{\sqrt{x+1}} - 2\cdot\frac{1}{1+\sqrt{x+1}}\cdot\frac{1}{2\sqrt{x+1}} = \frac{1}{\sqrt{x+1}} - \frac{1}{\sqrt{x+1}(1+\sqrt{x+1})} = \frac{1}{\sqrt{x+1}}\cdot\frac{(1+\sqrt{x+1})-1}{1+\sqrt{x+1}} = \frac{1}{1+\sqrt{x+1}}. $$(A1) Matches the integrand exactly.(A1) 与被积函数完全吻合。