Math Insight

Integration by parts


Strangely, the subtlest standard method is just the product rule run backwards. This is called integration by parts. (This might seem strange because often people find the chain rule for differentiation harder to get a grip on than the product rule). One way of writing the integration by parts rule is $$\int f(x)\cdot g'(x)\;dx=f(x)g(x)-\int f'(x)\cdot g(x)\;dx$$ Sometimes this is written another way: if we use the notation that for a function $u$ of $x$, $$du={du\over dx}\;dx$$ then for two functions $u,v$ of $x$ the rule is $$\int u\;dv=uv-\int v\;du.$$

Yes, it is hard to see how this might be helpful, but it is. The first theme we'll see in examples is where we could do the integral except that there is a power of $x$ ‘in the way’:

The simplest example is $$\int x\;e^x\;dx=\int x\;d(e^x)=x\,e^x-\int e^x\;dx=x\,e^x-e^x+C$$ Here we have taken $u=x$ and $v=e^x$. It is important to be able to see the $e^x$ as being the derivative of itself.

A similar example is $$\int x\;\cos x\;dx=\int x\;d(\sin x)= x\,\sin x-\int \sin x\;dx=x\,\sin x+\cos x +C$$ Here we have taken $u=x$ and $v=\sin x$. It is important to be able to see the $\cos x$ as being the derivative of $\sin x$.

Yet another example, illustrating also the idea of repeating the integration by parts: \begin{align*} \int x^2\;e^x\;dx&=\int x^2\;d(e^x)=x^2\,e^x-\int e^x\;d(x^2)\\ &=x^2\,e^x-2 \int x\,e^x\;dx=x^2\,e^x-2x\,e^x+2\int e^x\;dx\\ &=x^2\,e^x-2x\,e^x+2e^x+C \end{align*} Here we integrate by parts twice. After the first integration by parts, the integral we come up with is $\int xe^x\,dx$, which we had dealt with in the first example.

Or sometimes the theme is that it is easier to integrate the derivative of something than to integrate the thing: \begin{align*} \int \ln x\;dx&=\int \ln x\;d(x)=x\ln x-\int x\; d(\ln x)\\ &=x\ln x-\int x\;{1\over x}\; dx=x\ln x-\int 1\; dx=x\ln x-x+C \end{align*} We took $u=\ln x$ and $v=x$.

Again in this example it is easier to integrate the derivative than the thing itself: \begin{align*} \int \arctan x\;dx&=\int \arctan x\;d(x)=x\arctan x-\int x\; d(\arctan x)\\ &=x\arctan x-\int {x\over 1+x^2}\;dx =x\arctan x-{1\over 2}\int {2x\over 1+x^2}\;dx\\ &=x\arctan x-{1\over 2}\ln(1+x^2)+C \end{align*} since we should recognize the $${2x\over 1+x^2}$$ as being the derivative (via the chain rule) of $\ln(1+x^2)$.


  1. $\int \ln\,x\,dx=?$
  2. $\int xe^x\,dx=?$
  3. $\int (\ln\,x)^2\,dx=?$
  4. $\int xe^{2x}\,dx=?$
  5. $\int \arctan\,3x\,dx=?$
  6. $\int x^3\ln x\,dx=?$
  7. $\int \ln\,3x\,dx=?$
  8. $\int x\ln x\,dx=?$