Question

Solve each of the integrals in Exercises $39–74$. Some integrals require trigonometric substitution, and some do not. Write your answers as algebraic functions whenever possible.
$\int \ln \left(x^2+1\right)dx$

Short answer

The value is$x\ln \left(x^2+1\right)-2x+2{\tan }^{-1}x+C$

Step-by-step solution

Step 1. Given information

An expression is given as$\int \ln \left(x^2+1\right)dx$

Step 2. Solving Integral

Let

$$\begin{gathered} u=\ln \left(x^2+1\right) \\ du=\frac{2x}{x^2+1}dx \\ v=\int 1dx \\ =x \end{gathered}$$

Now substitute the value and simplify as,

$$\begin{gathered} \int \ln \left(x^2+1\right)dx \\ =\left(\ln \left(x^2+1\right)x-\int x\times \frac{2x}{x^2+1}dx\right) \\ =x\ln \left(x^2+1\right)-\int \frac{2x^2}{x^2+1}dx \\ =x\ln \left(x^2+1\right)-2\int \left(1-\frac{1}{x^2+1}\right)dx \\ =x\ln \left(x^2+1\right)-2x+2{\tan }^{-1}x+C \end{gathered}$$