Short and Sweet Calculus

5.4 Integration by Parts

A powerful technique of integration is integration by parts, which hangs on the formula for differential of a product. Let \(u\) and \(v\) be two functions of \(x\). Then \[d(uv)=udv+vdu\quad\Rightarrow\quad udv=d(uv)-vdu.\] Integrating both sides, we have \[\int udv=\int d(uv)+\int vdu.\] Since \(\int d(uv)=uv+C\), we get \[\boxed{\int udv=uv-\int vdu.\tag{a}}\] Because we have to add a constant when we integrate \(\int vdu\), we can ignore the constant of integration \(C\) when we integrate \(\int d(uv)\).

When we choose substitutions for \(u\) and \(v\), we usually want \(dv\) to be an expression that can be easily integrated and \(u\) to be a function that has a simple derivative.

Example 5.8. Evaluate \({\displaystyle \int x\sin xdx}\).


Let \[u=x\Rightarrow du=dx\] \[dv=\sin xdx\Rightarrow v=-\cos x+C_{1}\tag{*}\] \[\begin{aligned} \therefore\int x\sin xdx & =\underbrace{-x(\cos x+C_{1})}_{uv}-\int\underbrace{(-\cos x+C_{1})dx}_{vdu}\\ & =-x\cos x-C_{1}x+\sin x+C_{1}x+C\\ & =-x\cos x+\sin x+C.\end{aligned}\]

  • In the above example, we see that the first constant of integration \(C_{1}\) does not appear in the final result. We can prove that this is, in general, true and we do not need to write \(C_{1}\) when finding \(v\) from \(dv\). Suppose we write \(v+C_{1}\) in place of \(v\) in formula (a). Then we have \[\begin{aligned} \int udv & =u(v+C_{1})-\int(v+C_{1})du\\ & =uv+C_{1}u-C_{1}u-\int vdu\\ & =uv-\int vdu.\end{aligned}\] So in the following examples, we will not write the constant of the integral \(\int d(uv)\).

Example 5.9. Evaluate \({\displaystyle \int\ln xdx}.\)


Obviously we cannot put \(u=1\) and \(dv=\ln xdx\) because we do not know how to find \(v\) in this case. So we have to put \(u=\ln x\) and \(dv=dx\) \[u=\ln x\Rightarrow du=\frac{1}{x}dx\] \[dv=dx\Rightarrow v=x\] \[\begin{aligned} \therefore\int\ln xdx & =x\ln x-\int x\frac{1}{x}dx\\ & =x\ln x-x+C.\end{aligned}\]

Example 5.10. Evaluate \({\displaystyle \int x\ln xdx}.\)


Method 1: Let \(u=\ln x\) and \(dv=xdx\). Then \[u=\ln x\Rightarrow du=\frac{dx}{x},\] \[dv=xdx\Rightarrow v=\frac{1}{2}x^{2}.\] \[\begin{aligned} \therefore\int x\ln xdx & =\frac{1}{2}x^{2}\ln x-\int\frac{1}{2}x^{2}\frac{dx}{x}\\ & =\frac{1}{2}x^{2}\ln x-\frac{1}{4}x^{2}+C.\end{aligned}\] Method 2: Let \(u=x\) and \(dv=\ln xdx\). Then \[u=x\Rightarrow du=dx\] \[dv=\ln xdx\Rightarrow v=x\ln x-x\qquad\text{(from the previous example)}\] \[\begin{aligned} \therefore\int x\ln xdx & =x(x\ln x-x)-\int(x\ln x-x)dx\\ & =x^{2}\ln x-x^{2}-\int x\ln xdx+\int xdx\end{aligned}\] \[\Rightarrow2\int x\ln xdx=x^{2}\ln x-x^{2}+\frac{1}{2}x^{2}+C’\] Finally \[\int x\ln xdx=\frac{1}{2}x^{2}\ln x-\frac{1}{4}x^{2}+C’.\] Obviously the first method is easier, as in the second method we need to use integration by parts twice.

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