Question

Integrate:

$\int \frac{1+x^2}{1-x^2}dx$

• A. $I=\log \left(\frac{x+1}{x-1}\right)-x+C$
• B. $I=\log \left(\frac{x-1}{x+1}\right)-x+C$
• C. $I=\log \left(\frac{x+2}{x-1}\right)-x+C$
• D. $I=\log \left(\frac{x-2}{x-1}\right)-x+C$

Answer 1

Answer: (A)

$I=\log \left(\frac{x+1}{x-1}\right)-x+C$

We have,

$I=\int \frac{1+x^2}{1-x^2}dx$

$I=-\int \frac{1+x^2}{x^2-1}dx$

$I=-\int \frac{x^2+1-1+1}{x^2-1}dx$

$I=-\int \frac{x^2-1}{x^2-1}dx-\int \frac{2}{x^2-1}dx$

$I=-\int 1dx-2\int \frac{1}{x^2-1}dx$

$I=-x-2\left(\frac{1}{2}\log \left(\frac{x-1}{x+1}\right)\right)+C$

$I=-x+\log \left(\frac{x+1}{x-1}\right)+C$

$I=\log \left(\frac{x+1}{x-1}\right)-x+C$

Hence, this is the answer.