Integration by parts

1. When $u$ and $v$ change by small amounts, $uv$ gains two strips and a corner. After dividing by the change in $x$, the corner contribution tends to zero in the derivative limit.
2. That is the product rule: the blue strip is $v\,\frac{du}{dx}$, the orange strip is $u\,\frac{dv}{dx}$.
3. Integrate both sides: the whole rectangle is the two strip integrals added up.
4. Move the blue integral across. This rearrangement is integration by parts.

1. Differentiate $xe^{x}$ with the product rule: two strips, $e^{x}$ and $xe^{x}$.
2. Integrate both sides. The strip we cannot do directly, $\int xe^{x}\,dx$, is on the right.
3. Move the easy strip across: $\int xe^{x}\,dx = xe^{x}-e^{x}+c$.
4. Check it: differentiate the answer and the two $e^{x}$ terms cancel.

1. $\frac{d}{dx}(x\sin x)=\sin x+x\cos x$
2. $x\sin x=\int \sin x\,dx+\int x\cos x\,dx$
3. $\int x\cos x\,dx=x\sin x-\int \sin x\,dx$
4. $\int x\cos x\,dx=x\sin x+\cos x+c$

1. $\frac{d}{dx}(xe^{2x})=e^{2x}+2xe^{2x}$
2. $xe^{2x}=\int e^{2x}\,dx+2\int xe^{2x}\,dx$
3. $2\int xe^{2x}\,dx=xe^{2x}-\tfrac12 e^{2x}$
4. $\int xe^{2x}\,dx=\tfrac12 xe^{2x}-\tfrac14 e^{2x}+c$

1. Put one factor in as $u$, the other as $\frac{dv}{dx}$. Differentiate down the left, integrate down the right.
2. Read the formula off the frame: the diagonal is $uv$; the bottom row is the new integral.
3. Choose the other way round and the new integral has $x^{2}$ in it: worse than where you started.
4. So $u$ is the factor that gets simpler when you differentiate it.

1. $u=x$, because it differentiates to $1$. $\frac{dv}{dx}=\sin x$ integrates to $-\cos x$.
2. Fill the formula: $uv$ first, then minus the integral of $v$ times $\frac{du}{dx}$.
3. Two minuses make a plus: the last integral is $\int \cos x\,dx$.
4. $\int x\sin x\,dx=-x\cos x+\sin x+c$.

Tap the factor of $\displaystyle\int x\cos 2x\,dx$ that should be $u$.

1. ${u=x},\ {\frac{dv}{dx}=e^{3x}}\ \Rightarrow\ {\frac{du}{dx}=1},\ {v=\tfrac13 e^{3x}}$
2. $\int xe^{3x}\,dx=\tfrac13 xe^{3x}-\int \tfrac13 e^{3x}\,dx$
3. $=\tfrac13 xe^{3x}-\tfrac19 e^{3x}+c$

1. ${u=x},\ {\frac{dv}{dx}=\sin 3x}\ \Rightarrow\ {\frac{du}{dx}=1},\ {v=-\tfrac13\cos 3x}$
2. $\int x\sin 3x\,dx=-\tfrac13 x\cos 3x+\int \tfrac13\cos 3x\,dx$
3. $=-\tfrac13 x\cos 3x+\tfrac19\sin 3x+c$

1. ${u=3x+2},\ {\frac{dv}{dx}=e^{-x}}\ \Rightarrow\ {\frac{du}{dx}=3},\ {v=-e^{-x}}$
2. $\int {(3x+2)}e^{-x}\,dx=-{(3x+2)}e^{-x}+\int 3e^{-x}\,dx$
3. $=-{(3x+2)}e^{-x}-3e^{-x}+c$
4. $=-(3x+5)e^{-x}+c$

1. $u=x^{2}$. One round of parts leaves $\int xe^{x}\,dx$: still a product, but the power has dropped.
2. Go again on the new integral. Down the ladder: $x^{2}$, $2x$, $2$; the $e^{x}$ never changes.
3. Two rounds for $x^{2}$: one round for each power of $x$.
4. $\int x^{2}e^{x}\,dx=(x^{2}-2x+2)e^{x}+c$.

1. $u=x^{2}$, $\frac{dv}{dx}=\cos x$. First round: $x^{2}\sin x$ minus $\int 2x\sin x\,dx$.
2. Second round on $\int 2x\sin x\,dx$: $u=2x$, and $\sin x$ integrates to $-\cos x$.
3. Bracket the second round: the minus in front flips both its signs.
4. $\int x^{2}\cos x\,dx=x^{2}\sin x+2x\cos x-2\sin x+c$.

Tap the factor of $\displaystyle\int x^{2}\sin x\,dx$ that should be $u$.

1. ${u=x^{2}},\ {\frac{dv}{dx}=e^{2x}}\ \Rightarrow\ {\frac{du}{dx}=2x},\ {v=\tfrac12 e^{2x}}$
2. $\int x^{2}e^{2x}\,dx=\tfrac12 x^{2}e^{2x}-\int xe^{2x}\,dx$
3. $\int xe^{2x}\,dx=\tfrac12 xe^{2x}-\int \tfrac12 e^{2x}\,dx=\tfrac12 xe^{2x}-\tfrac14 e^{2x}$
4. $\int x^{2}e^{2x}\,dx=\tfrac12 x^{2}e^{2x}-\tfrac12 xe^{2x}+\tfrac14 e^{2x}+c$
5. $=\tfrac14\left(2x^{2}-2x+1\right)e^{2x}+c$

1. ${u=x^{2}},\ {\frac{dv}{dx}=\cos 2x}\ \Rightarrow\ {\frac{du}{dx}=2x},\ {v=\tfrac12\sin 2x}$
2. $\int x^{2}\cos 2x\,dx=\tfrac12 x^{2}\sin 2x-\int x\sin 2x\,dx$
3. $\int x\sin 2x\,dx=-\tfrac12 x\cos 2x+\int \tfrac12\cos 2x\,dx=-\tfrac12 x\cos 2x+\tfrac14\sin 2x$
4. $\int x^{2}\cos 2x\,dx=\tfrac12 x^{2}\sin 2x+\tfrac12 x\cos 2x-\tfrac14\sin 2x+c$

1. ${u=x^{2}-3},\ {\frac{dv}{dx}=e^{-x}}\ \Rightarrow\ {\frac{du}{dx}=2x},\ {v=-e^{-x}}$
2. $\int {(x^{2}-3)}e^{-x}\,dx=-{(x^{2}-3)}e^{-x}+\int 2xe^{-x}\,dx$
3. $\int 2xe^{-x}\,dx=-2xe^{-x}+\int 2e^{-x}\,dx=-2xe^{-x}-2e^{-x}$
4. $\int {(x^{2}-3)}e^{-x}\,dx=-{(x^{2}-3)}e^{-x}-2xe^{-x}-2e^{-x}+c$
5. $=-\left(x^{2}+2x-1\right)e^{-x}+c$

1. As $x$ runs from $a$ to $b$, the point $(u,v)$ slides up the curve.
2. Under the curve is $\int v\,\frac{du}{dx}\,dx$; beside it, up to the $v$-axis, is $\int u\,\frac{dv}{dx}\,dx$.
3. Together they fill an L: the big rectangle minus the small one, which is $[uv]$ between the limits.
4. Move the blue integral across. Every piece carries the limits $a$ and $b$.

1. Same parts, with limits: the $uv$ term becomes a bracket, evaluated straight away.
2. $[xe^{x}]_{0}^{1}=e$. The last integral, $\int_{0}^{1}e^{x}\,dx$, is $e-1$.
3. The area is exactly $1$: every $e$ cancelled.

Drag the upper limit $b$ so that the area under $y=xe^{x}$ from $0$ to $b$ is exactly $1$.

1. ${u=x},\ {\frac{dv}{dx}=\sin x}\ \Rightarrow\ {\frac{du}{dx}=1},\ {v=-\cos x}$
2. $\int_{0}^{\pi}x\sin x\,dx=\big[{-x\cos x}\big]_{0}^{\pi}+\int_{0}^{\pi}\cos x\,dx$
3. $\big[{-x\cos x}\big]_{0}^{\pi}=-\pi\cos\pi-0=\pi$
4. $\int_{0}^{\pi}\cos x\,dx=\big[{\sin x}\big]_{0}^{\pi}=0$
5. $\int_{0}^{\pi}x\sin x\,dx=\pi$

1. ${u=x},\ {\frac{dv}{dx}=e^{x}}\ \Rightarrow\ {\frac{du}{dx}=1},\ {v=e^{x}}$
2. $\int_{0}^{\ln 2}xe^{x}\,dx=\big[{xe^{x}}\big]_{0}^{\ln 2}-\int_{0}^{\ln 2}e^{x}\,dx$
3. $\big[{xe^{x}}\big]_{0}^{\ln 2}=(\ln 2)\,e^{\ln 2}-0=2\ln 2$
4. $\int_{0}^{\ln 2}e^{x}\,dx=\big[{e^{x}}\big]_{0}^{\ln 2}=2-1=1$
5. $\int_{0}^{\ln 2}xe^{x}\,dx=2\ln 2-1$

1. ${u=x},\ {\frac{dv}{dx}=\cos 2x}\ \Rightarrow\ {\frac{du}{dx}=1},\ {v=\tfrac12\sin 2x}$
2. $\int_{0}^{\pi/4}x\cos 2x\,dx=\big[{\tfrac12 x\sin 2x}\big]_{0}^{\pi/4}-\int_{0}^{\pi/4}\tfrac12\sin 2x\,dx$
3. $\big[{\tfrac12 x\sin 2x}\big]_{0}^{\pi/4}=\tfrac12\cdot\tfrac{\pi}{4}\cdot\sin\tfrac{\pi}{2}-0=\tfrac{\pi}{8}$
4. $\int_{0}^{\pi/4}\tfrac12\sin 2x\,dx=\big[{-\tfrac14\cos 2x}\big]_{0}^{\pi/4}=0-\left(-\tfrac14\right)=\tfrac14$
5. $\int_{0}^{\pi/4}x\cos 2x\,dx=\frac{\pi}{8}-\frac14$

1. To derive an integral of $\ln x$, write it as $1\times\ln x$ and use parts, with $x>0$.
2. $u=\ln x$: it differentiates to $\frac{1}{x}$. $\frac{dv}{dx}=1$ integrates to $x$.
3. $x$ times $\frac{1}{x}$ is $1$. The integral that is left is just $\int 1\,dx$.
4. $\int \ln x\,dx=x\ln x-x+c$.

1. For $x\ln x$, take $u=\ln x$: its derivative $1/x$ simplifies the remaining integral. Here $x>0$.
2. $v=\frac{x^{2}}{2}$, and $v$ times $\frac{1}{x}$ collapses to $\frac{x}{2}$.
3. $\int x\ln x\,dx=\frac{x^{2}}{2}\ln x-\frac{x^{2}}{4}+c$.

Tap the factor of $\displaystyle\int x^{2}\ln x\,dx$ that should be $u$.

1. ${u=\ln x},\ {\frac{dv}{dx}=x^{3}}\ \Rightarrow\ {\frac{du}{dx}=\frac{1}{x}},\ {v=\tfrac14 x^{4}}$
2. $\int x^{3}\ln x\,dx=\tfrac14 x^{4}\ln x-\int \tfrac14 x^{4}\cdot\frac{1}{x}\,dx=\tfrac14 x^{4}\ln x-\int \tfrac14 x^{3}\,dx$
3. $=\tfrac14 x^{4}\ln x-\tfrac{1}{16}x^{4}+c$

1. $\text{Area}=\int_{1}^{e}\ln x\,dx$ (the curve meets the axis at $x=1$)
2. ${u=\ln x},\ {\frac{dv}{dx}=1}\ \Rightarrow\ {\frac{du}{dx}=\frac{1}{x}},\ {v=x}$
3. $\int_{1}^{e}\ln x\,dx=\big[{x\ln x}\big]_{1}^{e}-\int_{1}^{e}1\,dx$
4. $\big[{x\ln x}\big]_{1}^{e}=e\cdot 1-1\cdot 0=e,\qquad \int_{1}^{e}1\,dx=e-1$
5. $\text{Area}=e-(e-1)=1$

1. ${u=\ln x},\ {\frac{dv}{dx}=x^{1/2}}\ \Rightarrow\ {\frac{du}{dx}=\frac{1}{x}},\ {v=\tfrac23 x^{3/2}}$
2. $\int_{1}^{4}\sqrt{x}\ln x\,dx=\big[{\tfrac23 x^{3/2}\ln x}\big]_{1}^{4}-\int_{1}^{4}\tfrac23 x^{1/2}\,dx$
3. $\big[{\tfrac23 x^{3/2}\ln x}\big]_{1}^{4}=\tfrac23\cdot 8\ln 4-0=\tfrac{16}{3}\ln 4$
4. $\int_{1}^{4}\tfrac23 x^{1/2}\,dx=\big[{\tfrac49 x^{3/2}}\big]_{1}^{4}=\tfrac49(8-1)=\tfrac{28}{9}$
5. $\tfrac{16}{3}\ln 4-\tfrac{28}{9}=\frac{32}{3}\ln 2-\frac{28}{9}$

1. ${u=x},\ {\frac{dv}{dx}=e^{-2x}}\ \Rightarrow\ {\frac{du}{dx}=1},\ {v=-\tfrac12 e^{-2x}}$
2. $\int xe^{-2x}\,dx=-\tfrac12 xe^{-2x}+\int \tfrac12 e^{-2x}\,dx$
3. $=-\tfrac12 xe^{-2x}-\tfrac14 e^{-2x}+c$
4. $=-\tfrac14\left(2x+1\right)e^{-2x}+c$

1. ${u=x},\ {\frac{dv}{dx}=\sec^{2}x}\ \Rightarrow\ {\frac{du}{dx}=1},\ {v=\tan x}$
2. $\int x\sec^{2}x\,dx=x\tan x-\int \tan x\,dx$
3. $\int \tan x\,dx=\ln\lvert\sec x\rvert+c$
4. $\int x\sec^{2}x\,dx=x\tan x-\ln\lvert\sec x\rvert+c$

1. ${u=x},\ {\frac{dv}{dx}=e^{2x}}\ \Rightarrow\ {\frac{du}{dx}=1},\ {v=\tfrac12 e^{2x}}$
2. $\int_{0}^{1/2}xe^{2x}\,dx=\big[{\tfrac12 xe^{2x}}\big]_{0}^{1/2}-\int_{0}^{1/2}\tfrac12 e^{2x}\,dx$
3. $\big[{\tfrac12 xe^{2x}}\big]_{0}^{1/2}=\tfrac12\cdot\tfrac12\cdot e-0=\tfrac14 e$
4. $\int_{0}^{1/2}\tfrac12 e^{2x}\,dx=\big[{\tfrac14 e^{2x}}\big]_{0}^{1/2}=\tfrac14 e-\tfrac14$
5. $\tfrac14 e-\left(\tfrac14 e-\tfrac14\right)=\frac14$

1. (a) $u=\ln x$: it differentiates to $\frac{1}{x}$, so the remaining integrand is the simpler power $x^4/5$.
2. (b) ${\frac{dv}{dx}=x^{4}}\ \Rightarrow\ {\frac{du}{dx}=\frac{1}{x}},\ {v=\tfrac15 x^{5}}$
3. $\int_{1}^{2}x^{4}\ln x\,dx=\big[{\tfrac15 x^{5}\ln x}\big]_{1}^{2}-\int_{1}^{2}\tfrac15 x^{4}\,dx$
4. $\big[{\tfrac15 x^{5}\ln x}\big]_{1}^{2}=\tfrac{32}{5}\ln 2-0=\tfrac{32}{5}\ln 2$
5. $\int_{1}^{2}\tfrac15 x^{4}\,dx=\big[{\tfrac{1}{25}x^{5}}\big]_{1}^{2}=\tfrac{32}{25}-\tfrac{1}{25}=\tfrac{31}{25}$
6. $\int_{1}^{2}x^{4}\ln x\,dx=\frac{32}{5}\ln 2-\frac{31}{25}$

1. ${u=(\ln x)^{2}},\ {\frac{dv}{dx}=1}\ \Rightarrow\ {\frac{du}{dx}=\frac{2\ln x}{x}},\ {v=x}$
2. $\int (\ln x)^{2}\,dx=x(\ln x)^{2}-\int 2\ln x\,dx$
3. $\int \ln x\,dx=x\ln x-\int x\cdot\frac{1}{x}\,dx$
4. $\int \ln x\,dx=x\ln x-x$
5. $\int (\ln x)^{2}\,dx=x(\ln x)^{2}-2x\ln x+2x+c$

1. $\text{Area}=\int_{0}^{2}xe^{-x}\,dx$
2. ${u=x},\ {\frac{dv}{dx}=e^{-x}}\ \Rightarrow\ {\frac{du}{dx}=1},\ {v=-e^{-x}}$
3. $\int_{0}^{2}xe^{-x}\,dx=\big[{-xe^{-x}}\big]_{0}^{2}+\int_{0}^{2}e^{-x}\,dx$
4. $\big[{-xe^{-x}}\big]_{0}^{2}=-2e^{-2}-0=-2e^{-2}$
5. $\int_{0}^{2}e^{-x}\,dx=\big[{-e^{-x}}\big]_{0}^{2}=-e^{-2}+1$
6. $\text{Area}=1-3e^{-2}$

1. ${u=x^{2}},\ {\frac{dv}{dx}=\sin x}\ \Rightarrow\ {\frac{du}{dx}=2x},\ {v=-\cos x}$
2. $\int_{0}^{\pi/2}x^{2}\sin x\,dx=\big[{-x^{2}\cos x}\big]_{0}^{\pi/2}+\int_{0}^{\pi/2}2x\cos x\,dx$
3. $\big[{-x^{2}\cos x}\big]_{0}^{\pi/2}=-\tfrac{\pi^{2}}{4}\cos\tfrac{\pi}{2}-0=0$
4. ${u=2x},\ {\frac{dv}{dx}=\cos x}\ \Rightarrow\ {\frac{du}{dx}=2},\ {v=\sin x}$
5. $\int_{0}^{\pi/2}2x\cos x\,dx=\big[{2x\sin x}\big]_{0}^{\pi/2}-\int_{0}^{\pi/2}2\sin x\,dx$
6. $=\pi-\big[{-2\cos x}\big]_{0}^{\pi/2}=\pi-(0+2)=\pi-2$
7. $\int_{0}^{\pi/2}x^{2}\sin x\,dx=0+{(\pi-2)}=\pi-2$