Integration by Parts Practice – 3 Levels

Integration by Parts Formula

$$\int u\,dv = uv – \int v\,du$$

Choose $u$ so that $du$ is simpler, and $dv$ so that $v$ is easy to find.

Tip: LIATE order for choosing $u$ — Log · Inverse trig · Algebraic · Trig · Exponential

Worked Example

Evaluate $\displaystyle\int x\,e^{x}\,dx$

1st: Choose $u$ and $dv$

$u = x$

$dv = e^{x}\,dx$

2nd: Find $du$ and $v$

$u = x$

$du = dx$

$v = e^{x}$

$dv = e^{x}\,dx$

3. Substitute

$uv – \int v\,du$

$x e^{x} – \int e^{x}\,dx$

4. Integrate & simplify

$x e^{x} – e^{x} + C$

$e^{x}(x – 1) + C$

Level 1

Easy

Straightforward products — one application of integration by parts

$\displaystyle\int x\cos x\,dx$

Show Answer

1st: Choose $u$ and $dv$

$u = x$

$dv = \cos x\,dx$

2nd: Find $du$ and $v$

$u = x$

$du = dx$

$v = \sin x$

$dv = \cos x\,dx$

3. Substitute

$uv – \int v\,du$

$x\sin x – \int\sin x\,dx$

4. Integrate & simplify

$x\sin x + \cos x + C$

$\displaystyle\int x e^{2x}\,dx$

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1st: Choose $u$ and $dv$

$u = x$

$dv = e^{2x}\,dx$

2nd: Find $du$ and $v$

$u = x$

$du = dx$

$v = \dfrac{1}{2}e^{2x}$

$dv = e^{2x}\,dx$

3. Substitute

$uv – \int v\,du$

$x \cdot \dfrac{1}{2}e^{2x} – \int \dfrac{1}{2}e^{2x}\,dx$

4. Integrate & simplify

$\dfrac{1}{2}x e^{2x} – \dfrac{1}{4}e^{2x} + C$

$\dfrac{e^{2x}}{4}(2x – 1) + C$

$\displaystyle\int\ln x\,dx$

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1st: Choose $u$ and $dv$

$u = \ln x$

$dv = dx$

2nd: Find $du$ and $v$

$u = \ln x$

$du = \dfrac{1}{x}\,dx$

$v = x$

$dv = dx$

3. Substitute

$uv – \int v\,du$

$x\ln x – \int x \cdot \dfrac{1}{x}\,dx$

4. Integrate & simplify

$x\ln x – \int 1\,dx = x\ln x – x + C$

$x\ln x – x + C$

Level 2

Medium

Thoughtful choice of $u$ / $dv$, or a substitution first

$\displaystyle\int x^{3} e^{x^{2}}\,dx$

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1st: Choose $u$ and $dv$

(Rewrite first: $\int x^{2}\cdot(x e^{x^{2}})\,dx$)

$u = x^{2}$

$dv = x e^{x^{2}}\,dx$

2nd: Find $du$ and $v$

$u = x^{2}$

$du = 2x\,dx$

$v = \dfrac{1}{2}e^{x^{2}}$

$dv = x e^{x^{2}}\,dx$

3. Substitute

$uv – \int v\,du$

$x^{2}\cdot\dfrac{1}{2}e^{x^{2}} – \int\dfrac{1}{2}e^{x^{2}}\cdot 2x\,dx$

4. Integrate & simplify

$\dfrac{1}{2}x^{2}e^{x^{2}} – \int x e^{x^{2}}\,dx = \dfrac{1}{2}x^{2}e^{x^{2}} – \dfrac{1}{2}e^{x^{2}} + C$

$\dfrac{1}{2}e^{x^{2}}(x^{2} – 1) + C$

$\displaystyle\int x^{2}\sin x\,dx$

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1st: Choose $u$ and $dv$

$u = x^{2}$

$dv = \sin x\,dx$

2nd: Find $du$ and $v$

$u = x^{2}$

$du = 2x\,dx$

$v = -\cos x$

$dv = \sin x\,dx$

3. Substitute

$uv – \int v\,du$

$-x^{2}\cos x – \int(-\cos x)(2x)\,dx = -x^{2}\cos x + 2\int x\cos x\,dx$

4. Integrate & simplify (apply IBP again to $\int x\cos x\,dx$)

For $\int x\cos x\,dx$: $u=x$, $dv=\cos x\,dx$ → $x\sin x + \cos x$

$-x^{2}\cos x + 2(x\sin x + \cos x) + C$

$-x^{2}\cos x + 2x\sin x + 2\cos x + C$

$\displaystyle\int\arctan x\,dx$

Show Answer

1st: Choose $u$ and $dv$

$u = \arctan x$

$dv = dx$

2nd: Find $du$ and $v$

$u = \arctan x$

$du = \dfrac{1}{1+x^{2}}\,dx$

$v = x$

$dv = dx$

3. Substitute

$uv – \int v\,du$

$x\arctan x – \int\dfrac{x}{1+x^{2}}\,dx$

4. Integrate & simplify

Let $w=1+x^{2}$ → $\dfrac{1}{2}\ln|1+x^{2}|$

$x\arctan x – \dfrac{1}{2}\ln(1+x^{2}) + C$

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