Integration by partsEdexcel International A Level Maths: Revision notes

Integration by parts reverses the product rule. Since $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$, rearranging and integrating gives $\int u\frac{dv}{dx}\,dx=uv-\int v\frac{du}{dx}\,dx.$ Use it when the integrand is a product of two different types of function that cannot be integrated by substitution. For a definite integral the $uv$ term is evaluated between the limits: $\int_a^bu\frac{dv}{dx}\,dx=\left[uv\right]_a^b-\int_a^bv\frac{du}{dx}\,dx$.

Choose $u$ so that differentiating it makes it simpler, and $\frac{dv}{dx}$ so that you can integrate it. A guide is to let $u$ be the first of: a logarithm, an algebraic term such as $x$ or $x^2$, a trigonometric function, an exponential. Example: $\int xe^{3x}\,dx$. Let $u=x$, $\frac{dv}{dx}=e^{3x}$, so $\frac{du}{dx}=1$ and $v=\frac13e^{3x}$: $\int xe^{3x}\,dx=\frac13xe^{3x}-\int\frac13e^{3x}\,dx=\frac13xe^{3x}-\frac19e^{3x}+c.$

There is no standard integral for $\ln x$, but write it as $1\times\ln x$. Let $u=\ln x$ and $\frac{dv}{dx}=1$, so $\frac{du}{dx}=\frac1x$ and $v=x$: $\int\ln x\,dx=x\ln x-\int x\cdot\frac1x\,dx=x\ln x-x+c.$ Example: the area under $y=\ln x$ from $x=1$ to $x=e$ is $\left[x\ln x-x\right]_1^e=(e-e)-(0-1)=1$. The same idea works for $\int x^n\ln x\,dx$ with $u=\ln x$.

If after one application the new integral still needs parts, apply the method again. Example: $I=\int_0^1x^2e^{x}\,dx$. With $u=x^2$, $\frac{dv}{dx}=e^{x}$: $I=\left[x^2e^{x}\right]_0^1-2\int_0^1xe^{x}\,dx=e-2\int_0^1xe^{x}\,dx$. For $\int xe^{x}\,dx$ use $u=x$: $xe^{x}-e^{x}$, so $\int_0^1xe^{x}\,dx=1$. Hence $I=e-2$. The power of $x$ falls by one each time, so $x^n$ needs $n$ applications.

For $\int e^{x}\sin x\,dx$ neither function becomes simpler, but two applications return the original integral $I$. First: $I=e^{x}\sin x-\int e^{x}\cos x\,dx$. Second: $\int e^{x}\cos x\,dx=e^{x}\cos x+\int e^{x}\sin x\,dx$. So $I=e^{x}\sin x-e^{x}\cos x-I$, which rearranges to $I=\tfrac12e^{x}(\sin x-\cos x)+c.$ Keep the same choice of $u$ (the trigonometric function) both times, otherwise the two applications cancel out. For areas, split the integral where the curve crosses the $x$-axis and add the positive areas.