Integration by partsAQA A-Level Maths: Revision notes

The product rule gives $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$. Integrating and rearranging: $\int u\frac{dv}{dx}\,dx=uv-\int v\frac{du}{dx}\,dx.$ This is integration by parts. It turns an integral that is a product into $uv$ plus a new integral, which is chosen to be easier. For a definite integral use $\left[uv\right]_a^b-\int_a^bv\frac{du}{dx}dx$, applying the limits to the $uv$ term as well.

Choose $u$ so that differentiating it makes it simpler, and $\frac{dv}{dx}$ so that it integrates easily. A good order for $u$ is: $\ln x$, then powers of $x$, then trigonometric, then exponential. For $\int xe^{2x}dx$ take $u=x$ (it becomes $1$) and $\frac{dv}{dx}=e^{2x}$ (so $v=\frac12e^{2x}$). The wrong choice $u=e^{2x}$, $\frac{dv}{dx}=x$ makes the power of $x$ rise to $\frac{x^2}{2}$ and the new integral is worse.

$\int xe^{2x}dx$: $u=x$, $v=\frac12e^{2x}$, so $=\frac x2e^{2x}-\int\frac12e^{2x}dx=\frac x2e^{2x}-\frac14e^{2x}+c=\frac{e^{2x}}{4}(2x-1)+c$. $\int x\cos x\,dx$: $u=x$, $v=\sin x$, so $=x\sin x-\int\sin x\,dx=x\sin x+\cos x+c$. Definite: $\int_0^{\pi/2}x\cos x\,dx=\left[x\sin x+\cos x\right]_0^{\pi/2}=\frac\pi2-1$. Always differentiate your answer to check it returns the integrand.

To integrate $\ln x$ write it as $\ln x\times1$ with $u=\ln x$, $\frac{dv}{dx}=1$. Then $\frac{du}{dx}=\frac1x$, $v=x$ and $\int\ln x\,dx=x\ln x-\int x\cdot\frac1x\,dx=x\ln x-x+c.$ Use this to find areas: $\int_1^{e^2}\ln x\,dx=\left[x\ln x-x\right]_1^{e^2}=(2e^2-e^2)-(0-1)=e^2+1$, using $\ln e^2=2$ and $\ln1=0$. For $\int x^2\ln x\,dx$ take $u=\ln x$, $\frac{dv}{dx}=x^2$, which gives $\frac{x^3}{3}\ln x-\frac{x^3}{9}+c$.

When the new integral still contains a product, apply the method again. For $\int x^2e^xdx$: $u=x^2$ gives $x^2e^x-\int2xe^xdx$, then $u=2x$ gives $\int2xe^xdx=2xe^x-2e^x$. So $\int x^2e^x\,dx=e^x(x^2-2x+2)+c,\qquad\int_0^1x^2e^x\,dx=e-2.$ The power of $x$ drops by one each time, so $x^n$ needs $n$ applications. Take care with the minus sign and the bracket in the second stage: subtract the whole second integral. Reduction formulae are not required.