Question
Integrate the rational function $\frac{x^{3}+x+1}{x^{2}-1}$

Answer

Given function is $\frac{x^{3}+x+1}{x^{2}-1}$ 
Dividing (x3 + x + 1) by x2 - 1, we get,
$\frac{x^{3}+x+1}{x^{2}-1}=x+\frac{2 x+1}{x^{2}-1}$ 
Let $\frac{2 x+1}{x^{2}-1}=\frac{A}{(x+1)}+\frac{B}{(x-1)}$ 
Now, 2x + 1 = A(x – 1) + B(x + 1) ...(i)
Substituting x = 1 and -1 in equation (i), we get,
$A=\frac{1}{2}$ and $B=\frac{3}{2}$ 
Thus, $\frac{x^{3}+x+1}{x^{2}-1}=x+\frac{1}{2(x+1)}+\frac{3}{2(x-1)}$
$\Rightarrow~~\int \frac{x^{3}+x+1}{x^{2}-1}=\int x d x+\frac{1}{2} \int \frac{1}{(x+1)} d x+\frac{3}{2} \int \frac{1}{(x-1)} d x$ 
= $\frac{x^{2}}{2}+\frac{1}{2} \log |x+1|+\frac{3}{2} \log |x-1|+C$  

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