Question

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

Answer 1

The given integrand is not a proper fraction.

Therefore, on dividing $(x^3+x+1)$ with $x^2-1$, we get,

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

Let $\frac{2x+1}{x^2-1}=\frac{A}{(x+1)}+\frac{B}{(x-1)}$

$2x+1=A(x-1)+B(x+1)$

On solving and equating the coefficients, we get,

$A=\frac{1}{2}$ and $B=\frac{3}{2}$

Therefore,

$\frac{x^3+x+1}{x^2-1}=x+\frac{1}{2(x+1)}+\frac{3}{2(x-1)}$

$\int \frac{x^3+x+1}{x^2-1}dx=\int xdx+\frac{1}{2}\int \frac{1}{(x+1)}dx+\frac{3}{2}\int \frac{1}{(x-1)}dx$

$=\frac{x^2}{2}+\frac{1}{2}\log |x+1|+\frac{3}{2}\log |x-1|+C$