Use a Table of Integrals to find $\int x^{2}\tan ^{-1}x\,dx$.

Solution Find the subsection of the table titled Integrals Containing Inverse Trigonometric Functions. Then look for an integral whose form closely resembles the problem. You should find Integral 114: $$\int x^{n}\,\tan ^{-1}x\,dx=\dfrac{1}{n+1} \left( x^{n+1}\,\tan ^{-1}x-\int \dfrac{x^{n+1}\,dx}{1+x^{2}}\right)\qquad n\neq -1$$

521

This is the integral we seek with $n=2.$ $$\int x^{2} \tan ^{-1}x~dx = \dfrac{1}{3}\left( x^{3} \tan^{-1}x-\int \dfrac{x^{3} dx}{1+x^{2}}\right)$$

We find the integral on the right by using the substitution $u=1+x^{2}.$ Then $du=2x\,dx$ and $$\begin{eqnarray*} \int \dfrac{x^{3}}{1+x^{2}}dx &=& \int \dfrac{x^{2}x\,dx}{1+x^{2}}=\int \dfrac{u-1 }{u}\dfrac{du}{2}=\dfrac{1}{2}\int \left( 1-\dfrac{1}{u}\right) du=\dfrac{1}{ 2}u-\dfrac{1}{2}\ln \vert u \vert \\[6pt] &=& \dfrac{1+x^{2}}{2}-\dfrac{\ln (1+x^{2}) }{2} \end{eqnarray*}$$

So, $$\begin{eqnarray*} \int x^{2}\,\tan ^{-1}x\,dx &=& \dfrac{1}{3}\left( x^{3}\,\tan ^{-1}x-\int \dfrac{x^{3}}{1+x^{2}}\,dx\right) \\[6pt] &=& \dfrac{1}{3}\left[ x^{3}\,\tan ^{-1}x- \dfrac{1+x^{2}}{2}+\dfrac{1}{2}\ln ( 1+x^{2}) \right] +C \end{eqnarray*}$$