Integration by Parts

Integration by parts is the integral counterpart of the product rule for differentiation, and it is the go-to method whenever the integrand is a product of two functions of different types — or a single function that has no direct integration formula of its own, such as $\log x$ or $\sin^{-1}x$.

If $u$ and $v$ are differentiable functions of $x$, the product rule gives $d(uv)=v\,du+u\,dv$, i.e. $u\,dv=d(uv)-v\,du$. Integrating both sides,

This is the integration-by-parts formula:

It doesn't hand back a finished answer directly — it trades the original integral $\int u\,dv$ for a different integral $\int v\,du$, which is only useful if that second integral is easier than the first. (This partial nature of the result is why some countries call the technique "partial integration" rather than "integration by parts" — both names describe the same rule.)

Choosing $u$ and $dv$ correctly is the whole game, and the book gives three guiding rules, applied in order:

1. if the integrand contains a function that has no direct integration formula — $\log x$, $\tan^{-1}x$, $\sin^{-1}x$, etc. — that function must be taken as $u$, with everything else as $dv$;
2. if the integrand is a product of two directly-integrable functions and one of them is $x^n$ ($n$ a positive integer), take $u=x^n$ (this is what lets the power of $x$ eventually disappear after repeated differentiation);
3. in any other case, the choice of $u$ is free — pick whichever assignment makes $\int v\,du$ simpler than the original. Worked-style illustrations (paralleling Example 11.33). For $\int xe^x\,dx$: since $x$ is algebraic and $e^x$ exponential, rule (ii) says take $u=x$ (so $du=dx$) and $dv=e^x\,dx$ (so $v=e^x$); then $\int xe^x\,dx=xe^x-\int e^x\,dx=xe^x-e^x+c$. For $\int\log x\,dx$: $\log x$ has no direct formula, so by rule (i), $u=\log x$ ($du=\tfrac1x dx$) and $dv=dx$ ($v=x$), giving $\int\log x\,dx=x\log x-\int x\cdot\tfrac1x\,dx=x\log x-x+c$. The same idea handles $\int\sin^{-1}x\,dx$: take $u=\sin^{-1}x$, $dv=dx$; then $v=x$, $du=\dfrac{dx}{\sqrt{1-x^2}}$, and $\int\sin^{-1}x\,dx=x\sin^{-1}x-\displaystyle\int\dfrac{x\,dx}{\sqrt{1-x^2}}$, and the remaining integral falls to a quick substitution $t=1-x^2$, giving finally $x\sin^{-1}x+\sqrt{1-x^2}+c$. …