Companion to the University-Style Practice Set配套大学风格练习题集
Sections 1 to 7: substitution (indefinite and definite), integration by parts, repeated and tabular IBP, the cyclic trick, reduction formulas, and method selection第 1 至 7 节：换元法（不定积分与定积分）、分部积分法、多次及列表分部积分、循环积分技巧、递推公式与方法选择CALC II
Evaluate: (a) $\int x^{2}\sin(x^{3}+1)\,dx$; (b) $\int_{0}^{1}\frac{2x}{(x^{2}+1)^{3}}\,dx$; (c) $\int_{1}^{e^{2}}\frac{(\ln x)^{2}}{x}\,dx$.计算：(a) $\int x^{2}\sin(x^{3}+1)\,dx$；(b) $\int_{0}^{1}\frac{2x}{(x^{2}+1)^{3}}\,dx$；(c) $\int_{1}^{e^{2}}\frac{(\ln x)^{2}}{x}\,dx$。
The integrand is of the form $f'(g(x))\cdot g'(x)$ with $g(x)=x^{3}+1$. (M1) Set $u=x^{3}+1$, so $du=3x^{2}\,dx$ and $x^{2}\,dx=\tfrac{du}{3}$:被积函数具有 $f'(g(x))\cdot g'(x)$ 的形式，其中 $g(x)=x^{3}+1$。(M1) 令 $u=x^{3}+1$，则 $du=3x^{2}\,dx$，$x^{2}\,dx=\tfrac{du}{3}$：
$$ \int x^{2}\sin(x^{3}+1)\,dx = \int \sin u\cdot\frac{du}{3} = -\frac{1}{3}\cos u+C = -\frac{1}{3}\cos(x^{3}+1)+C. $$Verification: differentiating gives $-\tfrac{1}{3}\cdot(-\sin(x^{3}+1))\cdot 3x^{2}=x^{2}\sin(x^{3}+1)$. (A1)验证：求导得 $-\tfrac{1}{3}\cdot(-\sin(x^{3}+1))\cdot 3x^{2}=x^{2}\sin(x^{3}+1)$。(A1)
Set $u=x^{2}+1$, $du=2x\,dx$. (M1) Limits: $x=0\Rightarrow u=1$; $x=1\Rightarrow u=2$. (A1)令 $u=x^{2}+1$，$du=2x\,dx$。(M1) 积分限：$x=0\Rightarrow u=1$；$x=1\Rightarrow u=2$。(A1)
$$ \int_{0}^{1}\frac{2x}{(x^{2}+1)^{3}}\,dx = \int_{1}^{2}u^{-3}\,du = \left[-\frac{1}{2u^{2}}\right]_{1}^{2} = -\frac{1}{8}+\frac{1}{2} = \frac{3}{8}. $$(A1)
Set $u=\ln x$, $du=\tfrac{dx}{x}$. (M1) Limits: $x=1\Rightarrow u=0$; $x=e^{2}\Rightarrow u=2$. (A1)令 $u=\ln x$，$du=\tfrac{dx}{x}$。(M1) 积分限：$x=1\Rightarrow u=0$；$x=e^{2}\Rightarrow u=2$。(A1)
$$ \int_{1}^{e^{2}}\frac{(\ln x)^{2}}{x}\,dx = \int_{0}^{2}u^{2}\,du = \left[\frac{u^{3}}{3}\right]_{0}^{2} = \frac{8}{3}. $$(A1)
Evaluate using integration by parts once: (a) $\int x\,e^{2x}\,dx$; (b) $\int\ln x\,dx$.各使用一次分部积分法计算：(a) $\int x\,e^{2x}\,dx$；(b) $\int\ln x\,dx$。
By LIATE, the algebraic factor $x$ is differentiated and the exponential is integrated: $u=x$, $dv=e^{2x}\,dx$, so $du=dx$ and $v=\tfrac{1}{2}e^{2x}$. (M1)按 LIATE，代数因子 $x$ 求导，指数函数积分：$u=x$，$dv=e^{2x}\,dx$，则 $du=dx$，$v=\tfrac{1}{2}e^{2x}$。(M1)
$$ \int x\,e^{2x}\,dx = \frac{x}{2}e^{2x} - \int\frac{1}{2}e^{2x}\,dx = \frac{x}{2}e^{2x} - \frac{1}{4}e^{2x}+C = \frac{e^{2x}}{4}(2x-1)+C. $$(A1) Verification: $\tfrac{d}{dx}\!\left[\tfrac{e^{2x}}{4}(2x-1)\right]=\tfrac{2e^{2x}}{4}(2x-1)+\tfrac{e^{2x}}{4}\cdot 2=\tfrac{e^{2x}}{2}(2x-1)+\tfrac{e^{2x}}{2}=xe^{2x}$. (A1)(A1) 验证：$\tfrac{d}{dx}\!\left[\tfrac{e^{2x}}{4}(2x-1)\right]=\tfrac{2e^{2x}}{4}(2x-1)+\tfrac{e^{2x}}{4}\cdot 2=\tfrac{e^{2x}}{2}(2x-1)+\tfrac{e^{2x}}{2}=xe^{2x}$。(A1)
There is no second factor, so write $\int\ln x\cdot 1\,dx$. LIATE gives $u=\ln x$, $dv=dx$, so $du=\tfrac{1}{x}\,dx$ and $v=x$. (M1)没有第二个因子，将其写成 $\int\ln x\cdot 1\,dx$。LIATE 给出 $u=\ln x$，$dv=dx$，则 $du=\tfrac{1}{x}\,dx$，$v=x$。(M1)
$$ \int\ln x\,dx = x\ln x - \int x\cdot\frac{1}{x}\,dx = x\ln x - \int 1\,dx = x\ln x - x + C. $$(A1) Verification: $\ln x + 1 - 1 = \ln x$. (A1)(A1) 验证：$\ln x + 1 - 1 = \ln x$。(A1)
Evaluate $I=\int x^{3}e^{x}\,dx$ by (a) applying IBP three times; (b) constructing the tabular table and confirming.计算 $I=\int x^{3}e^{x}\,dx$：(a) 连续使用三次分部积分法；(b) 构造列表并验证。
Step 1.第 1 步。 $u=x^{3}$, $dv=e^{x}\,dx$, $du=3x^{2}\,dx$, $v=e^{x}$. (M1)
$$ I = x^{3}e^{x} - 3\int x^{2}e^{x}\,dx. $$Step 2.第 2 步。 Apply IBP to $\int x^{2}e^{x}\,dx$:对 $\int x^{2}e^{x}\,dx$ 进行分部积分： $u=x^{2}$, $dv=e^{x}\,dx$, $du=2x\,dx$, $v=e^{x}$. (M1)
$$ \int x^{2}e^{x}\,dx = x^{2}e^{x}-2\int xe^{x}\,dx. $$Step 3.第 3 步。 Apply IBP to $\int xe^{x}\,dx$:对 $\int xe^{x}\,dx$ 进行分部积分： $u=x$, $dv=e^{x}\,dx$, $du=dx$, $v=e^{x}$. (M1)
$$ \int xe^{x}\,dx = xe^{x}-e^{x}. $$Assembling from the innermost out: (A1)从最内层向外组合：(A1)
$$ I = x^{3}e^{x} - 3\!\left[x^{2}e^{x}-2(xe^{x}-e^{x})\right] = x^{3}e^{x}-3x^{2}e^{x}+6xe^{x}-6e^{x}+C = e^{x}(x^{3}-3x^{2}+6x-6)+C. $$Verification: $\tfrac{d}{dx}[e^{x}(x^{3}-3x^{2}+6x-6)]=e^{x}(x^{3}-3x^{2}+6x-6)+e^{x}(3x^{2}-6x+6)=e^{x}\cdot x^{3}$. (A1)验证：$\tfrac{d}{dx}[e^{x}(x^{3}-3x^{2}+6x-6)]=e^{x}(x^{3}-3x^{2}+6x-6)+e^{x}(3x^{2}-6x+6)=e^{x}\cdot x^{3}$。(A1)
Draw two columns: differentiate $x^{3}$ successively on the left, integrate $e^{x}$ successively on the right, alternating signs $+,-,+,-,+$. (M1)画两列：左列对 $x^{3}$ 逐次求导，右列对 $e^{x}$ 逐次积分，符号交替取 $+,-,+,-,+$。(M1)
| Sign符号 | Differentiate ($u$-column)求导列（$u$ 列） | Integrate ($dv$-column)积分列（$dv$ 列） |
|---|---|---|
| $+$ | $x^{3}$ | $e^{x}$ |
| $-$ | $3x^{2}$ | $e^{x}$ |
| $+$ | $6x$ | $e^{x}$ |
| $-$ | $6$ | $e^{x}$ |
| $+$ | $0$ | $e^{x}$ |
Multiply diagonally (each left entry times the entry one row down on the right), applying the sign: (A1)沿对角线相乘（左列每项乘以右列下一行对应项），并施加相应符号：(A1)
$$ I = (+1)\cdot x^{3}e^{x} + (-1)\cdot 3x^{2}e^{x} + (+1)\cdot 6xe^{x} + (-1)\cdot 6e^{x} + C = e^{x}(x^{3}-3x^{2}+6x-6)+C. $$This matches part (a). (A1)与 (a) 的结果一致。(A1)
Let $I=\int e^{x}\sin x\,dx$. (a) Apply IBP twice, keeping $e^x$ as the $dv$ factor each time, and show $I$ reappears on the right. (b) Hence solve for $I$.设 $I=\int e^{x}\sin x\,dx$。(a) 进行两次分部积分，每次均以 $e^x$ 为 $dv$ 因子，证明 $I$ 重新出现在等式右侧。(b) 由此解出 $I$。
Choose $u=\sin x$, $dv=e^{x}\,dx$, so $du=\cos x\,dx$, $v=e^{x}$. (M1)取 $u=\sin x$，$dv=e^{x}\,dx$，则 $du=\cos x\,dx$，$v=e^{x}$。(M1)
$$ I = e^{x}\sin x - \int e^{x}\cos x\,dx. $$(A1)
Apply IBP to $\int e^{x}\cos x\,dx$, again keeping $e^{x}$ in the $dv$ role: $u=\cos x$, $dv=e^{x}\,dx$, so $du=-\sin x\,dx$, $v=e^{x}$. (M1)对 $\int e^{x}\cos x\,dx$ 进行分部积分，仍以 $e^{x}$ 担任 $dv$：$u=\cos x$，$dv=e^{x}\,dx$，则 $du=-\sin x\,dx$，$v=e^{x}$。(M1)
$$ \int e^{x}\cos x\,dx = e^{x}\cos x - \int e^{x}(-\sin x)\,dx = e^{x}\cos x + \int e^{x}\sin x\,dx = e^{x}\cos x + I. $$Substituting back: $I = e^{x}\sin x - (e^{x}\cos x + I) = e^{x}\sin x - e^{x}\cos x - I$. (A1) The integral $I$ has returned on the right-hand side.代回得：$I = e^{x}\sin x - (e^{x}\cos x + I) = e^{x}\sin x - e^{x}\cos x - I$。(A1) 积分 $I$ 重新出现在等式右侧。
Rearrange: $I + I = e^{x}\sin x - e^{x}\cos x$, so (M1)整理：$I + I = e^{x}\sin x - e^{x}\cos x$，从而 (M1)
$$ 2I = e^{x}(\sin x - \cos x), \qquad I = \frac{e^{x}(\sin x - \cos x)}{2}+C. $$(A1) Verification: $\tfrac{d}{dx}\!\left[\tfrac{e^{x}(\sin x-\cos x)}{2}\right]=\tfrac{e^{x}(\sin x-\cos x)+e^{x}(\cos x+\sin x)}{2}=\tfrac{2e^{x}\sin x}{2}=e^{x}\sin x$.(A1) 验证：$\tfrac{d}{dx}\!\left[\tfrac{e^{x}(\sin x-\cos x)}{2}\right]=\tfrac{e^{x}(\sin x-\cos x)+e^{x}(\cos x+\sin x)}{2}=\tfrac{2e^{x}\sin x}{2}=e^{x}\sin x$。
(a) Derive $\int u\,dv=uv-\int v\,du$ from the product rule. (b) Derive $I_{n}=x^{n}e^{x}-nI_{n-1}$. (c) Evaluate $I_{3}$ using the reduction formula with $I_{0}=e^{x}+C$.(a) 从乘积法则推导 $\int u\,dv=uv-\int v\,du$。(b) 推导 $I_{n}=x^{n}e^{x}-nI_{n-1}$。(c) 利用递推公式及 $I_{0}=e^{x}+C$ 计算 $I_{3}$。
Let $u=u(x)$ and $v=v(x)$ be differentiable. The product rule states $(uv)'=u'v+uv'$. (M1) Integrate both sides with respect to $x$:设 $u=u(x)$ 与 $v=v(x)$ 均可微。乘积法则给出 $(uv)'=u'v+uv'$。(M1) 对两边关于 $x$ 积分：
$$ uv = \int u'v\,dx + \int uv'\,dx. $$Rearranging: $\int uv'\,dx = uv - \int u'v\,dx$. Writing $dv=v'(x)\,dx$ and $du=u'(x)\,dx$ gives the IBP formula (A1)整理得：$\int uv'\,dx = uv - \int u'v\,dx$。记 $dv=v'(x)\,dx$，$du=u'(x)\,dx$，即得分部积分公式 (A1)
$$ \int u\,dv = uv - \int v\,du. $$The formula is valid whenever $u$ and $v$ are differentiable and the integrals exist. (R1)只要 $u$ 与 $v$ 可微且积分存在，该公式即成立。(R1)
Let $I_{n}=\int x^{n}e^{x}\,dx$. Set $u=x^{n}$, $dv=e^{x}\,dx$, so $du=nx^{n-1}\,dx$ and $v=e^{x}$. (M1) Applying the IBP formula:设 $I_{n}=\int x^{n}e^{x}\,dx$。令 $u=x^{n}$，$dv=e^{x}\,dx$，则 $du=nx^{n-1}\,dx$，$v=e^{x}$。(M1) 应用分部积分公式：
$$ I_{n} = x^{n}e^{x} - \int e^{x}\cdot nx^{n-1}\,dx. $$(A1) The remaining integral is $n\int x^{n-1}e^{x}\,dx = nI_{n-1}$. (A1) Therefore(A1) 余下的积分为 $n\int x^{n-1}e^{x}\,dx = nI_{n-1}$。(A1) 因此
$$ I_{n} = x^{n}e^{x} - n\,I_{n-1}. \qquad \text{(A1)} $$Start from $I_{0}=e^{x}+C$. (M1)从 $I_{0}=e^{x}+C$ 出发。(M1)
$$ I_{1} = xe^{x}-1\cdot I_{0} = xe^{x}-e^{x}+C = e^{x}(x-1)+C. $$ $$ I_{2} = x^{2}e^{x}-2I_{1} = x^{2}e^{x}-2e^{x}(x-1)+C = e^{x}(x^{2}-2x+2)+C. \quad\text{(A1)} $$ $$ I_{3} = x^{3}e^{x}-3I_{2} = x^{3}e^{x}-3e^{x}(x^{2}-2x+2)+C = e^{x}(x^{3}-3x^{2}+6x-6)+C. \quad\text{(A1)} $$This matches the direct calculation in Q3.此结果与 Q3 的直接计算一致。
$I_{n}=\int_{0}^{\pi/2}\sin^{n}x\,dx$. (a) Derive $I_{n}=\tfrac{n-1}{n}I_{n-2}$. (b) Evaluate $I_{1}$. (c) Compute $I_{5}$.$I_{n}=\int_{0}^{\pi/2}\sin^{n}x\,dx$。(a) 推导 $I_{n}=\tfrac{n-1}{n}I_{n-2}$。(b) 计算 $I_{1}$。(c) 计算 $I_{5}$。
Write $\sin^{n}x=\sin^{n-1}x\cdot\sin x$ and apply IBP: set $u=\sin^{n-1}x$, $dv=\sin x\,dx$. Then $du=(n-1)\sin^{n-2}x\cos x\,dx$ and $v=-\cos x$. (M1)将 $\sin^{n}x$ 写成 $\sin^{n-1}x\cdot\sin x$，进行分部积分：令 $u=\sin^{n-1}x$，$dv=\sin x\,dx$，则 $du=(n-1)\sin^{n-2}x\cos x\,dx$，$v=-\cos x$。(M1)
$$ I_{n} = \left[-\cos x\sin^{n-1}x\right]_{0}^{\pi/2} + (n-1)\int_{0}^{\pi/2}\cos^{2}x\sin^{n-2}x\,dx. $$(M1) Evaluate the boundary term: at $x=\tfrac{\pi}{2}$, $\cos(\tfrac{\pi}{2})=0$; at $x=0$, $\sin^{n-1}(0)=0$ for $n\ge 2$. Both endpoints contribute zero, so the boundary term vanishes. (A1)(M1) 计算边界项：在 $x=\tfrac{\pi}{2}$ 处，$\cos(\tfrac{\pi}{2})=0$；在 $x=0$ 处，当 $n\ge 2$ 时 $\sin^{n-1}(0)=0$。两端点均为零，边界项消失。(A1)
Use the Pythagorean identity $\cos^{2}x=1-\sin^{2}x$: (A1)利用勾股恒等式 $\cos^{2}x=1-\sin^{2}x$：(A1)
$$ I_{n} = (n-1)\int_{0}^{\pi/2}(1-\sin^{2}x)\sin^{n-2}x\,dx = (n-1)\!\left[I_{n-2}-I_{n}\right]. $$Expand the right side and collect $I_{n}$ terms: (A1)展开右侧并合并含 $I_{n}$ 的项：(A1)
$$ I_{n} + (n-1)I_{n} = (n-1)I_{n-2} \implies n\,I_{n} = (n-1)I_{n-2}. $$ $$ \boxed{I_{n} = \frac{n-1}{n}\,I_{n-2}.} \qquad\text{(A1)} $$Step down from $n=5$ using odd-index base $I_{1}=1$. (M1)从 $n=5$ 向下递推，利用奇数指标的初始值 $I_{1}=1$。(M1)
$$ I_{3} = \frac{2}{3}I_{1} = \frac{2}{3}. \qquad\text{(A1)} $$ $$ I_{5} = \frac{4}{5}I_{3} = \frac{4}{5}\cdot\frac{2}{3} = \frac{8}{15}. \qquad\text{(A1)} $$$J_{n}=\int x^{n}\ln x\,dx$, $x>0$. (a) Derive a closed form using IBP. (b) Verify by differentiation.$J_{n}=\int x^{n}\ln x\,dx$，$x>0$。(a) 利用分部积分法推导闭合形式。(b) 通过求导验证。
LIATE places $\ln x$ above any power, so set $u=\ln x$, $dv=x^{n}\,dx$. Then $du=\tfrac{1}{x}\,dx$ and $v=\tfrac{x^{n+1}}{n+1}$. (M1)LIATE 中对数函数优先于幂函数，故令 $u=\ln x$，$dv=x^{n}\,dx$，则 $du=\tfrac{1}{x}\,dx$，$v=\tfrac{x^{n+1}}{n+1}$。(M1)
$$ J_{n} = \frac{x^{n+1}\ln x}{n+1} - \int\frac{x^{n+1}}{n+1}\cdot\frac{1}{x}\,dx = \frac{x^{n+1}\ln x}{n+1} - \frac{1}{n+1}\int x^{n}\,dx. $$(A1·A1) Evaluating the remaining power integral:(A1·A1) 计算余下的幂函数积分：
$$ J_{n} = \frac{x^{n+1}\ln x}{n+1} - \frac{1}{n+1}\cdot\frac{x^{n+1}}{n+1}+C = \frac{x^{n+1}\ln x}{n+1}-\frac{x^{n+1}}{(n+1)^{2}}+C. \qquad\text{(A1)} $$Differentiate term by term using the product rule on $x^{n+1}\ln x$: (M1)逐项求导，对 $x^{n+1}\ln x$ 使用乘积法则：(M1)
$$ \frac{d}{dx}\left[\frac{x^{n+1}\ln x}{n+1}-\frac{x^{n+1}}{(n+1)^{2}}\right] = \frac{(n+1)x^{n}\ln x + x^{n}}{n+1} - \frac{(n+1)x^{n}}{(n+1)^{2}} = x^{n}\ln x + \frac{x^{n}}{n+1} - \frac{x^{n}}{n+1} = x^{n}\ln x. $$(A1) The two extra terms cancel exactly, confirming the antiderivative.(A1) 两个多余项恰好抵消，从而确认原函数正确。
Evaluate $\int_{0}^{1}x^{3}\sqrt{x^{2}+1}\,dx$ by letting $u=x^{2}+1$, then expanding and integrating.令 $u=x^{2}+1$，展开后积分，计算 $\int_{0}^{1}x^{3}\sqrt{x^{2}+1}\,dx$。
Let $u=x^{2}+1$. Then $du=2x\,dx$, so $x\,dx=\tfrac{du}{2}$, and also $x^{2}=u-1$. (M1) New limits: $x=0\Rightarrow u=1$; $x=1\Rightarrow u=2$. (A1)令 $u=x^{2}+1$，则 $du=2x\,dx$，$x\,dx=\tfrac{du}{2}$，且 $x^{2}=u-1$。(M1) 新积分限：$x=0\Rightarrow u=1$；$x=1\Rightarrow u=2$。(A1)
Write $x^{3}\,dx=x^{2}\cdot(x\,dx)=(u-1)\tfrac{du}{2}$ and $\sqrt{x^{2}+1}=\sqrt{u}$:将 $x^{3}\,dx=x^{2}\cdot(x\,dx)=(u-1)\tfrac{du}{2}$，$\sqrt{x^{2}+1}=\sqrt{u}$：
$$ \int_{0}^{1}x^{3}\sqrt{x^{2}+1}\,dx = \int_{1}^{2}(u-1)\sqrt{u}\cdot\frac{du}{2} = \frac{1}{2}\int_{1}^{2}(u-1)\sqrt{u}\,du. \qquad\text{(A1)} $$$(u-1)\sqrt{u}=u^{3/2}-u^{1/2}$. (M1)$(u-1)\sqrt{u}=u^{3/2}-u^{1/2}$。(M1)
$$ \frac{1}{2}\int_{1}^{2}\!\left(u^{3/2}-u^{1/2}\right)du = \frac{1}{2}\left[\frac{2u^{5/2}}{5}-\frac{2u^{3/2}}{3}\right]_{1}^{2}. $$At $u=2$: $u^{5/2}=4\sqrt{2}$ and $u^{3/2}=2\sqrt{2}$. (A1)在 $u=2$ 处：$u^{5/2}=4\sqrt{2}$，$u^{3/2}=2\sqrt{2}$。(A1)
At $u=1$: $u^{5/2}=1$ and $u^{3/2}=1$. Evaluate:在 $u=1$ 处：$u^{5/2}=1$，$u^{3/2}=1$。计算：
$$ = \frac{1}{2}\!\left[\left(\frac{8\sqrt{2}}{5}-\frac{4\sqrt{2}}{3}\right)-\left(\frac{2}{5}-\frac{2}{3}\right)\right]. $$Combine over a common denominator of 15: (A1)通分，公分母为 15：(A1)
$$ \frac{8\sqrt{2}}{5}-\frac{4\sqrt{2}}{3}=\sqrt{2}\left(\frac{24-20}{15}\right)=\frac{4\sqrt{2}}{15}; \qquad \frac{2}{5}-\frac{2}{3}=\frac{6-10}{15}=-\frac{4}{15}. $$(A1) No further simplification is possible, so the exact answer is $\dfrac{2(\sqrt{2}+1)}{15}$. (A1)(A1) 无法进一步化简，精确答案为 $\dfrac{2(\sqrt{2}+1)}{15}$。(A1)
$f(x)=x\ln x$ on $[1,e]$. (a) Find $f'(x)$ and monotonicity. (b) Evaluate $\int_{1}^{e}x\ln x\,dx$. (c) Find the exact area of the region bounded by $y=x\ln x$, the $x$-axis, and $x=1$, $x=e$.$f(x)=x\ln x$ 定义在 $[1,e]$ 上。(a) 求 $f'(x)$ 及单调性。(b) 计算 $\int_{1}^{e}x\ln x\,dx$。(c) 求由 $y=x\ln x$、$x$ 轴及直线 $x=1$、$x=e$ 围成区域的精确面积。
By the product rule, $f'(x)=\ln x + x\cdot\tfrac{1}{x}=\ln x+1$. (M1) For $x\in(1,e)$, $\ln x>0$, so $f'(x)>1>0$: $f$ is strictly increasing on $(1,e)$. (A1)由乘积法则，$f'(x)=\ln x + x\cdot\tfrac{1}{x}=\ln x+1$。(M1) 对 $x\in(1,e)$，$\ln x>0$，故 $f'(x)>1>0$：$f$ 在 $(1,e)$ 上严格单调递增。(A1)
Set $u=\ln x$, $dv=x\,dx$, so $du=\tfrac{1}{x}\,dx$ and $v=\tfrac{x^{2}}{2}$. (M1)令 $u=\ln x$，$dv=x\,dx$，则 $du=\tfrac{1}{x}\,dx$，$v=\tfrac{x^{2}}{2}$。(M1)
$$ \int_{1}^{e}x\ln x\,dx = \left[\frac{x^{2}\ln x}{2}\right]_{1}^{e} - \int_{1}^{e}\frac{x^{2}}{2}\cdot\frac{1}{x}\,dx = \left[\frac{x^{2}\ln x}{2}\right]_{1}^{e} - \int_{1}^{e}\frac{x}{2}\,dx. $$(A1) Evaluate the boundary term: at $x=e$, $\tfrac{e^{2}\ln e}{2}=\tfrac{e^{2}}{2}$; at $x=1$, $\tfrac{1\cdot 0}{2}=0$. (M1) So the boundary term is $\tfrac{e^{2}}{2}$.(A1) 计算边界项：$x=e$ 时，$\tfrac{e^{2}\ln e}{2}=\tfrac{e^{2}}{2}$；$x=1$ 时，$\tfrac{1\cdot 0}{2}=0$。(M1) 故边界项为 $\tfrac{e^{2}}{2}$。
$$ \int_{1}^{e}\frac{x}{2}\,dx = \left[\frac{x^{2}}{4}\right]_{1}^{e} = \frac{e^{2}}{4}-\frac{1}{4} = \frac{e^{2}-1}{4}. \qquad\text{(A1)} $$ $$ \int_{1}^{e}x\ln x\,dx = \frac{e^{2}}{2}-\frac{e^{2}-1}{4} = \frac{2e^{2}}{4}-\frac{e^{2}-1}{4} = \frac{e^{2}+1}{4}. \qquad\text{(A1)} $$On $[1,e]$: $f(1)=1\cdot\ln 1=0$ and $f$ is increasing with $f'(x)>0$, so $f(x)\ge f(1)=0$ throughout $[1,e]$. (M1) The curve lies on or above the $x$-axis, so the area equals the integral without absolute values. (A1)在 $[1,e]$ 上：$f(1)=1\cdot\ln 1=0$，且 $f$ 单调递增，$f'(x)>0$，故在整个 $[1,e]$ 上 $f(x)\ge f(1)=0$。(M1) 曲线在 $x$ 轴上方（或与之重合），面积等于积分值，无需取绝对值。(A1)
$$ \text{Area} = \int_{1}^{e}x\ln x\,dx = \frac{e^{2}+1}{4}. \qquad\text{(A1)} $$For each integral, name the method, state the key step, then evaluate: (a) $\int\frac{x}{\sqrt{1-x^{2}}}\,dx$; (b) $\int x^{2}\cos x\,dx$; (c) $\int e^{x}\cos x\,dx$.对每个积分，说明方法、关键步骤，然后求值：(a) $\int\frac{x}{\sqrt{1-x^{2}}}\,dx$；(b) $\int x^{2}\cos x\,dx$；(c) $\int e^{x}\cos x\,dx$。
Method: substitution. Key step: $x\,dx$ is proportional to $d(1-x^{2})$. (M1) Set $u=1-x^{2}$, $du=-2x\,dx$, so $x\,dx=-\tfrac{du}{2}$:方法：换元法。关键步骤：$x\,dx$ 与 $d(1-x^{2})$ 成比例。(M1) 令 $u=1-x^{2}$，$du=-2x\,dx$，故 $x\,dx=-\tfrac{du}{2}$：
$$ \int\frac{x}{\sqrt{1-x^{2}}}\,dx = \int\frac{-du/2}{\sqrt{u}} = -\int\frac{1}{2}u^{-1/2}\,du = -u^{1/2}+C = -\sqrt{1-x^{2}}+C. $$(A1) Verification: $\tfrac{d}{dx}[-\sqrt{1-x^{2}}]=\tfrac{x}{\sqrt{1-x^{2}}}$.(A1) 验证：$\tfrac{d}{dx}[-\sqrt{1-x^{2}}]=\tfrac{x}{\sqrt{1-x^{2}}}$。
Method: tabular IBP. Key step: differentiate $x^{2}$ until it vanishes (two steps); integrate $\cos x$ correspondingly. (M1)方法：列表分部积分法。关键步骤：对 $x^{2}$ 逐次求导直至为零（两步）；对 $\cos x$ 逐次积分。(M1)
| Sign符号 | Diff.求导列 | Integ.积分列 |
|---|---|---|
| $+$ | $x^{2}$ | $\cos x$ |
| $-$ | $2x$ | $\sin x$ |
| $+$ | $2$ | $-\cos x$ |
| $-$ | $0$ | $-\sin x$ |
Diagonal products with alternating signs: (M1)沿对角线相乘并施加交替符号：(M1)
$$ \int x^{2}\cos x\,dx = x^{2}\sin x - 2x(-\cos x) + 2(-\sin x) + C = x^{2}\sin x + 2x\cos x - 2\sin x + C. $$(A1) Verification: $\tfrac{d}{dx}[x^{2}\sin x+2x\cos x-2\sin x]=2x\sin x+x^{2}\cos x+2\cos x-2x\sin x-2\cos x=x^{2}\cos x$. (A1)(A1) 验证：$\tfrac{d}{dx}[x^{2}\sin x+2x\cos x-2\sin x]=2x\sin x+x^{2}\cos x+2\cos x-2x\sin x-2\cos x=x^{2}\cos x$。(A1)
Method: cyclic IBP. Key step: two applications reproduce the original integral. Set $u=\cos x$, $dv=e^{x}\,dx$. (M1)方法：循环分部积分法。关键步骤：两次分部积分后还原原积分。令 $u=\cos x$，$dv=e^{x}\,dx$。(M1)
$$ \int e^{x}\cos x\,dx = e^{x}\cos x + \int e^{x}\sin x\,dx. $$(A1) Apply IBP to $\int e^{x}\sin x\,dx$ with $u=\sin x$, $dv=e^{x}\,dx$:(A1) 对 $\int e^{x}\sin x\,dx$ 取 $u=\sin x$，$dv=e^{x}\,dx$ 进行分部积分：
$$ \int e^{x}\sin x\,dx = e^{x}\sin x - \int e^{x}\cos x\,dx. $$(M1) Let $K=\int e^{x}\cos x\,dx$. Substituting back: $K=e^{x}\cos x+e^{x}\sin x-K$, so (A1)(M1) 设 $K=\int e^{x}\cos x\,dx$，代回得：$K=e^{x}\cos x+e^{x}\sin x-K$，从而 (A1)
$$ 2K = e^{x}(\cos x+\sin x), \qquad K = \frac{e^{x}(\cos x+\sin x)}{2}+C. $$Verification: $\tfrac{d}{dx}\!\left[\tfrac{e^{x}(\cos x+\sin x)}{2}\right]=\tfrac{e^{x}(\cos x+\sin x)+e^{x}(-\sin x+\cos x)}{2}=\tfrac{2e^{x}\cos x}{2}=e^{x}\cos x$.验证：$\tfrac{d}{dx}\!\left[\tfrac{e^{x}(\cos x+\sin x)}{2}\right]=\tfrac{e^{x}(\cos x+\sin x)+e^{x}(-\sin x+\cos x)}{2}=\tfrac{2e^{x}\cos x}{2}=e^{x}\cos x$。