Question

The value of the integral $\int \frac{x^3+x+1}{x^2-1}dx$ is
(where $C$ is an arbitrary constant)

• A. $\frac{x^2}{2}+\ln |x+1|+\frac{3}{2}\ln |x-1|+C$
• B. $\frac{x^2}{2}-\ln |x+1|+\frac{2}{3}\ln |x-1|+C$
• C. $\frac{x^2}{2}+\frac{1}{2}\ln |x+1|+\frac{3}{2}\ln |x-1|+C$
• D. $\frac{x^2}{2}+\ln |x-1|+\frac{3}{2}\ln |x+1|+C$

Answer 1

The correct option is C $\frac{x^2}{2}+\frac{1}{2}\ln |x+1|+\frac{3}{2}\ln |x-1|+C$

It can be seen that the given integrand is not a proper rational expression, therefore, on dividing
$\frac{x^3+x+1}{x^2-1}=x+\frac{2x+1}{x^2-1}$
Let
$\frac{2x+1}{x^2-1}=\frac{A}{(x+1)}+\frac{B}{(x-1)}\Rightarrow 2x+1=A(x-1)+B(x+1)\cdots \left(1\right)$
Equating the coefficients of $x$ and constant, we obtain
$A+B=2-A+B=1$
On solving, we obtain
$A=\frac{1}{2}$ and $B=\frac{3}{2}$

$\therefore \frac{x^3+x+1}{x^2-1}=x+\frac{1}{2(x+1)}+\frac{3}{2(x-1)}\Rightarrow \int \frac{x^3+x+1}{x^2+1}dx=\int xdx+\frac{1}{2}\int \frac{1}{(x+1)}dx+\frac{3}{2}\int \frac{1}{(x-1)}dx=\frac{x^2}{2}+\frac{1}{2}\ln |x+1|+\frac{3}{2}\ln |x-1|+C$