Integration by partsEdexcel A-Level Maths: Revision notes

Integration by parts is the reverse of the product rule. From $\frac{\mathrm{d}}{\mathrm{d}x}(uv)=u\frac{\mathrm{d}v}{\mathrm{d}x}+v\frac{\mathrm{d}u}{\mathrm{d}x}$, integrating both sides and rearranging gives $\int u\frac{\mathrm{d}v}{\mathrm{d}x}\,\mathrm{d}x=uv-\int v\frac{\mathrm{d}u}{\mathrm{d}x}\,\mathrm{d}x.$ Use it for a product of two different types of function, such as $xe^{2x}$, $x\sin x$ or $x\ln x$, that cannot be integrated by a simple rule or substitution. For a definite integral, evaluate $uv$ between the limits as well: $\left[uv\right]_a^b-\int_a^bv\frac{\mathrm{d}u}{\mathrm{d}x}\,\mathrm{d}x$.

Choose $u$ to be the part that becomes simpler when differentiated, and $\frac{\mathrm{d}v}{\mathrm{d}x}$ to be the part you can integrate. Take $u=\ln x$ if it appears; otherwise take $u$ to be the power of $x$; trigonometric and exponential functions are usually $\frac{\mathrm{d}v}{\mathrm{d}x}$. Example: $\int xe^{2x}\,\mathrm{d}x$ with $u=x$ and $\frac{\mathrm{d}v}{\mathrm{d}x}=e^{2x}$, so $\frac{\mathrm{d}u}{\mathrm{d}x}=1$, $v=\frac12e^{2x}$: $\int xe^{2x}\,\mathrm{d}x=\frac12xe^{2x}-\int\frac12e^{2x}\,\mathrm{d}x=\frac14e^{2x}(2x-1)+c.$

For $\int_0^1xe^{2x}\,\mathrm{d}x$, use the result above: $\left[\frac14e^{2x}(2x-1)\right]_0^1=\frac{e^2+1}{4}$. Another example: $\int_1^3\frac{\ln x}{x^2}\mathrm{d}x$ with $u=\ln x$ and $\frac{\mathrm{d}v}{\mathrm{d}x}=x^{-2}$, $v=-\frac1x$, gives $\left[-\frac{\ln x}{x}-\frac1x\right]_1^3=\frac{2-\ln3}{3}$. Check the answer is sensible by comparing with a rough numerical estimate of the area.

The integral of $\ln x$ is required knowledge. Write $\ln x=1\times\ln x$ with $u=\ln x$ and $\frac{\mathrm{d}v}{\mathrm{d}x}=1$, so $\frac{\mathrm{d}u}{\mathrm{d}x}=\frac1x$ and $v=x$: $\int\ln x\,\mathrm{d}x=x\ln x-\int1\,\mathrm{d}x=x\ln x-x+c.$ Then $\int_1^e\ln x\,\mathrm{d}x=\left[x\ln x-x\right]_1^e=1$. For $x^n\ln x$, use $u=\ln x$ and $\frac{\mathrm{d}v}{\mathrm{d}x}=x^n$: $\int x^2\ln x\,\mathrm{d}x=\frac{x^3}{3}\ln x-\frac{x^3}{9}+c$.

When the power of $x$ is 2 or more, one application leaves an integral that still needs parts. Each application lowers the power by one. Example: $\int x^2e^{3x}\,\mathrm{d}x=\frac{x^2e^{3x}}{3}-\frac23\int xe^{3x}\,\mathrm{d}x$, and a second application gives $\int xe^{3x}\,\mathrm{d}x=\frac{xe^{3x}}{3}-\frac{e^{3x}}{9}$. So the answer is $\frac{e^{3x}}{27}\left(9x^2-6x+2\right)+c$. Similarly $\int_0^{\pi/2}x^2\cos x\,\mathrm{d}x=\frac{\pi^2}{4}-2\int_0^{\pi/2}x\sin x\,\mathrm{d}x=\frac{\pi^2}{4}-2$. Reduction formulae are not needed.