Question

Solve:

$\int \frac{x^3+x+1}{x^2-1}dx$

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

Answer 1

The correct option is A

$\frac{x^2}{2}+\log (x^2-1)+\frac{1}{2}\log \frac{x-1}{x+1}+c$

Let $I=\int \frac{x^3+x+1}{x^2-1}dx$

Dividing the numerator by the denominator we get

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

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

$=\frac{x^2}{2}+\log (x^2-1)+\frac{1}{2}\log \frac{x-1}{x+1}+c$