Question

Solve$\int \frac{x}{4+x^2}dx$the following two ways:

(a) with the substitution $u=4+x^2;$

(b) with the trigonometric substitution x = 2 tan u.

Short answer

Part (a) The solution of the integral is $\frac{1}{2}\ln \left|4+x^2\right|+C.$

Part (b) The solution of the integral is $-\ln \left|\frac{2}{\sqrt{4+{\mathrm{x}}^2}}\right|+\mathrm{C}.$

Step-by-step solution

Part (a) Step 1. Given Information.

The given integral is$\int \frac{x}{4+x^2}dx.$

Part (a) Step 2. Solve. 

We have to solve the integral with the substitution $u=4+x^2,$so the derivative is$du=\left(2x\right)dx.$

Let's solve the integral by substituting u,

$$\begin{gathered} \int \frac{x}{4+x^2}dx \\ =\int \frac{x}{u}\left(\frac{du}{2x}\right) \\ =\frac{1}{2}\int \frac{du}{u} \\ =\frac{1}{2}\left[\ln \left|u\right|\right]+C \\ \mathrm{Substitue}\mathrm{back}u, \\ =\frac{1}{2}\ln \left|4+x^2\right|+C \end{gathered}$$

Part (b) Step 1. Solve. 

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

Let's solve the integral by substituting x,

$$\begin{gathered} \int \frac{x}{4+x^2}dx \\ =\int \frac{2\tan u}{4+{\left(2\tan u\right)}^2}\left(2se{c}^{2}u\right)du \\ =\int \frac{4\tan use{c}^{2}u}{4+\left(4{\tan }^{2}u\right)}du \\ =\int \frac{4\tan use{c}^{2}u}{4\left(1+{\tan }^{2}u\right)}du \\ =\int \frac{\tan use{c}^{2}u}{\left(1+{\tan }^{2}u\right)}du \\ \mathrm{Use}\mathrm{the}\mathrm{identity}1+{\tan }^2\mathrm{x}={\sec }^2\mathrm{x} \\ =\int \frac{\mathrm{tanu}{\sec }^2\mathrm{u}}{\left({\sec }^2\mathrm{u}\right)}\mathrm{du} \\ =\int \mathrm{tanu}\mathrm{du} \\ =-\ln \left|\mathrm{cosu}\right|+\mathrm{C} \\ \mathrm{Substitute}\mathrm{back}u, \\ =-\ln \left|\cos \left({\tan }^{-1}\left(\frac{\mathrm{x}}{2}\right)\right)\right|+\mathrm{C} \\ =-\ln \left|\frac{2}{\sqrt{4+{\mathrm{x}}^2}}\right|+\mathrm{C} \end{gathered}$$