Question

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

• A. $=\frac{x^2}{2}+log|x^2-1|+\frac{1}{2}log\left|\frac{x-1}{x+1}\right|+C$
• B. $=\frac{x^3}{2}+log|x^3-1|+\frac{1}{2}log\left|\frac{x-1}{x+1}\right|+C$
• C. $=\frac{3x^2}{2}+log|3x^2-1|+\frac{1}{2}log\left|\frac{x-1}{x+1}\right|+C$
• D. $=\frac{x^2}{5}+log|x^2+1|+\frac{1}{3}log\left|\frac{x-1}{x+1}\right|+C$

Answer 1

The correct option is A $=\frac{x^2}{2}+log|x^2-1|+\frac{1}{2}log\left|\frac{x-1}{x+1}\right|+C$

$\int \frac{x^3+x+1}{x^2-1}dx=\int (x+\frac{2x+1}{x^2-1})dx$
$=\int xdx+\int \frac{2x}{x^2-1}dx+\int \frac{1}{x^2-1}dx$
Put $x^2-1=t$ in the second integral
$=\frac{x^2}{2}+\int \frac{1}{t}dt+\frac{1}{2}log\left|\frac{x-1}{x+1}\right|+C$
$=\frac{x^2}{2}+log|x^2-1|+\frac{1}{2}log\left|\frac{x-1}{x+1}\right|+C$