Question

Solve$\int \frac{x^2}{1+x^2}dx$ the following two ways:

(a) with the substitution $u={\tan }^{-1}x;$

(b) with the trigonometric substitution x = tan u.

Short answer

Part (a) The solution of the given integral is $x-{\tan }^{-1}x+C.$

Part (b) The solution of the given integral is $x-{\tan }^{-1}x+C.$

Step-by-step solution

Part (a) Step 1. Given Information.

The given integral is$\int \frac{x^2}{1+x^2}dx.$

Part (a) Step 2. Solve.

We have to solve the integral with the substitution $u={\tan }^{-1}x,$so the derivative is $du=\frac{1}{1+x^2}dx.$

Let's solve the integral by substituting u,

$$\begin{gathered} \int \frac{x^2}{1+x^2}dx \\ =\int \frac{{\left(\tan u\right)}^2}{1+{\left(\tan u\right)}^2}\left(1+{\left(\tan u\right)}^2\right)du \\ =\int {\tan }^{2}udu \\ =\int -1+se{c}^{2}udu \\ =-\int 1du+\int se{c}^{2}udu \\ =-u+\tan u+C \\ \mathrm{Substitute}\mathrm{back}u \\ =-{\tan }^{-1}x+x+C \\ =x-{\tan }^{-1}x+C \end{gathered}$$

Part (b) Step 1. Solve.

We have to solve the integral with the substitution $x=\tan u,$so the derivative is $dx=se{c}^{2}udu.$

Let's solve the integral by substituting x,

role="math" localid="1648810032401" $$\begin{gathered} \int \frac{x^2}{1+x^2}dx \\ =\int \frac{{\tan }^{2}u}{1+{\tan }^{2}u}se{c}^{2}udu \\ =\int \frac{{\tan }^{2}u}{se{c}^{2}u}se{c}^{2}udu\left[1+{\tan }^{2}u=se{c}^{2}u\right] \\ =\int {\tan }^{2}udu \\ =\int -1+se{c}^{2}udu \\ =-\int 1du+\int se{c}^{2}udu \\ =-u+\tan u+C \\ \mathrm{Substitute}\mathrm{back}u \\ =-{\tan }^{-1}x+x+C \\ =x-{\tan }^{-1}x+C \end{gathered}$$