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

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Solution

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

$\begin{matrix} =^{⎛ ⎜ ⎝ \frac{3 (x^{2} - 1)}{{(x^{2} - 1)}^{3}} + \frac{4}{{(x^{2} - 1)}^{3}} ⎞ ⎟ ⎠} d x =^{⎛ ⎜ ⎝ \frac{3}{{(x^{2} - 1)}^{2}} + \frac{4}{{(x^{2} - 1)}^{3}} ⎞ ⎟ ⎠} d x \end{matrix}$

$= 3 \int^{\frac{1}{{(x^{2} - 1)}^{2}}} d x + 4 \int^{\frac{1}{{(x^{2} - 1)}^{3}}} d x$

$= \int^{(\frac{1}{4 (x + 1)} + \frac{1}{4 {(x + 1)}^{2}} - \frac{1}{4 (x - 1)} + \frac{1}{4 {(x - 1)}^{2}})} d x$

$= \frac{1}{4} \int^{\frac{1}{x + 1}} d x + \frac{1}{4} \int^{\frac{1}{{(x + 1)}^{2}}} d x - \frac{1}{4} \int^{\frac{1}{x - 1}} d x + \frac{1}{4} \int^{\frac{1}{{(x - 1)}^{2}}} d x$

$= \frac{1}{4} \int^{\frac{1}{x + 1}} d x + \frac{1}{4} \int^{\frac{1}{{(x + 1)}^{2}}} d x - \frac{1}{4} \int^{\frac{1}{x - 1}} d x + \frac{1}{4} \int^{\frac{1}{{(x - 1)}^{2}}} d x$

$= \frac{ln (x + 1)}{4} - \frac{1}{4 (x + 1)} - \frac{1}{4 (x - 1)} - \frac{ln (x - 1)}{4}$

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