Question

Integrate the rational functions.
$\int \frac{x^3+x+1}{x^2-1}dx.$

Answer 1

$\int \frac{x^3+x+1}{x^2-1}dx=\int xdx+\int \frac{2x+1}{x^2-1}dx[\because \frac{x^3+x+1}{x^2-1}=x+\frac{2x+1}{x^2-1}]=\int xdx+\int \frac{2x+1}{(x+1)(x-1)}dx$
Let $\frac{2x+1}{(x+1)(x-1)}=\frac{A}{(x+1)}+\frac{B}{(x-1)}$
$\Rightarrow \frac{2x+1}{(x+1)(x-1)}=\frac{A(x-1)+B(x+1)}{(x+1)(x-1)}\Rightarrow 2x+1=x(A+B)-A+B$
On comparing the coefficients of x and constant of x and constant term on both sides we get
$A+B=2$ and $-A+B=1$
On Adding above equations, we get $2B=3\Rightarrow B=\frac{3}{2}\text{and then}A=\frac{1}{2}$
$\therefore \int \frac{x^2+x+1}{x^2-1}dx=\int xdx+\int \frac{A}{(x+1)}dx+\int \frac{B}{(x-1)}dx=\int xdx+\frac{1}{2}\int \frac{1}{(x+1)}dx+\frac{3}{2}\int \frac{1}{(x-1)}dx=\frac{x^2}{2}+\frac{1}{2}log|x+1|+\frac{3}{2}log|x-1|+C$