# Interesting ways to evaluate integrals

#### Prove It

##### Well-known member
MHB Math Helper
Hi everyone. Just for fun I thought we could post some of the more interesting ways we know of to evaluate integrals

For starters, to evaluate \displaystyle \displaystyle \begin{align*} \int{\arctan{(x)}\,dx} \end{align*}, first we consider the integral \displaystyle \displaystyle \begin{align*} \int{\frac{2x}{1 + x^2}\,dx} \end{align*}. For simplicity, we'll leave out integration constants til the end...

We can integrate this using a substitution \displaystyle \displaystyle \begin{align*} u = 1 + x^2 \implies du = 2x\,dx \end{align*} and the integral becomes

\displaystyle \displaystyle \begin{align*} \int{\frac{2x}{1 + x^2}\,dx} &= \int{\frac{1}{u}\,du} \\ &= \ln{|u|} + C \\ &= \ln{ \left| 1 + x^2 \right| } + C \\ &= \ln{ \left( 1 + x^2 \right) } \textrm{ since } 1 + x^2 > 0 \textrm{ for all } x \in \mathbf{R} \end{align*}

Now supposing we wanted to evaluate the integral in a different way, using integration by parts with \displaystyle \displaystyle \begin{align*} u = 2x \implies du = 2\,dx \end{align*} and \displaystyle \displaystyle \begin{align*} dv = \frac{1}{1 + x^2}\,dx \implies v = \arctan{(x)} \end{align*}, then we would have

\displaystyle \displaystyle \begin{align*} \int{\frac{2x}{1 + x^2}\,dx} &= \int{2x \left( \frac{1}{1 + x^2} \right) dx} \\ &= 2x\arctan{(x)} - \int{2\arctan{(x)}\,dx} \\ &= 2x\arctan{(x)} - 2\int{\arctan{(x)}\,dx} \end{align*}

Now equating these gives

\displaystyle \displaystyle \begin{align*} \ln{ \left( 1 + x^2 \right) } &= 2x\arctan{(x)} - 2\int{\arctan{(x)}\,dx} \\ 2\int{\arctan{(x)}\,dx} &= 2x\arctan{(x)} - \ln{ \left( 1 + x^2 \right) } \\ \int{\arctan{(x)}\,dx} &= x\arctan{(x)} - \frac{1}{2}\ln{\left( 1 + x^2 \right) } + C \end{align*}

Q.E.D.

Staff member