Question

The value of $\int \left(\frac{x^2+1}{x^2-1}\right)·dx$ is

• A. $\log \left[\frac{(x–1)}{x+1}\right]+c$
• B. $\log \left[\frac{(x+1)}{x-1}\right]+c$
• C. $x+\log \left[\frac{(x–1)}{x+1}\right]+c$
• D. $\log \left[{(x-1)}^2\right]+c$

Answer 1

The correct option is C

$x+\log \left[\frac{(x–1)}{x+1}\right]+c$

Explanation for the correct option:

Compute the required value,

Given: $\int \left(\frac{x^2+1}{x^2-1}\right)·dx$

$$\begin{gathered} \int \left(\frac{x^2-1+2}{x^2-1}\right)·dx \\ =\int \left(\frac{x^2-1}{x^2-1}\right)·dx+\int \left(\frac{2}{x^2-1}\right)·dx \\ =\int \left(1\right)·dx+\int \left(\frac{2}{x^2-1}\right)·dx \\ =x+c+\int \left(\frac{x+1-(x-1)}{x^2-1}\right)·dx \\ =x+c+\int \left(\frac{x+1-(x-1)}{(x-1)(x+1)}\right)·dx \\ =x+c+\int \left(\frac{1}{(x-1)}\right)·dx-\int \left(\frac{1}{(x+1)}\right)·dx \\ =x+c_1+\log (x-1)+c_2-\log (x+1)+c_3\left[\int \frac{1}{x}·dx=\log \left(x\right)\right] \\ =x+\log \left(\frac{x-1}{x+1}\right)+C \end{gathered}$$

where $c_1,c_2,c_3,C$ are constants

Hence, option $C$ is the correct answer.