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Question:

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

Solution:

It can be seen that the given integrand is not a proper fraction.

Therefore, on dividing $\left(x^{3}+x+1\right)$ by $x^{2}-1$, we obtain

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

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

$2 x+1=A(x-1)+B(x+1)$    ...(1)

Substituting = 1 and −1 in equation (1), we obtain

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

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

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

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