Question:
Evaluate the following integral:
$\int \frac{x^{3}+x+1}{x^{2}-1}$
Solution:
Let
$I=\int \frac{x^{3}+x+1}{x^{2}-1} d x=\int\left(x+\frac{2 x+1}{x^{2}-1}\right) d x$
Now,
Let $\frac{2 x+1}{x^{2}-1}=\frac{A}{x+1}+\frac{B}{x-1}$
$2 x+1=A(x-1)+B(x+1)$
Put $x=1$
$2+1=A \times 0+B \times 2$
$3=2 B$
$B=\frac{3}{2}$
Put $x=-1$
$-2+1=-2 A+B \times 0$
$-1=-2 \mathrm{~A}$
$A=\frac{1}{2}$
$I=\int x d x+\frac{1}{2} \int \frac{d x}{x+1}+\frac{3}{2} \int \frac{d x}{x-1}$
$\int \frac{\mathrm{dx}}{\mathrm{x}}=\log |\mathrm{x}|$ and $\int \mathrm{xdx}=\frac{\mathrm{x}^{2}}{2}$
Therefore,
$I=\frac{x^{2}}{2}+\frac{1}{2} \log |x+1|+\frac{3}{2} \log |x-1|+c$
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