# How do you evaluate the integral int (3x+6)/(x^4-2x^2+1)?

Mar 24, 2018

$\frac{3}{2} L n \left(\frac{x + 1}{x - 1}\right) - \frac{6 x + 3}{2 {x}^{2} - 2} + C$

#### Explanation:

$\int \frac{3 x + 6}{{x}^{4} - 2 {x}^{2} + 1} \cdot \mathrm{dx}$

=$\int \frac{3 x + 6}{{x}^{2} - 1} ^ 2 \cdot \mathrm{dx}$

=$\int \frac{3 x + 6}{\left(x + 1\right) \cdot \left(x - 1\right)} ^ 2 \cdot \mathrm{dx}$

Now I decomposed integrand into basic fractions,

$\frac{3 x + 6}{\left(x + 1\right) \cdot \left(x - 1\right)} ^ 2 = \frac{A}{x + 1} + \frac{B}{x + 1} ^ 2 + \frac{C}{x - 1} + \frac{D}{x - 1} ^ 2$

After expanding denominator,

$A \left(x + 1\right) {\left(x - 1\right)}^{2} + B {\left(x - 1\right)}^{2} + C \left(x - 1\right) {\left(x + 1\right)}^{2} + D {\left(x + 1\right)}^{2} = 3 x + 6$ or

$A \cdot \left({x}^{3} - {x}^{2} - x + 1\right) + B \cdot \left({x}^{2} - 2 x + 1\right) + C \cdot \left({x}^{3} + {x}^{2} - x - 1\right) + D \cdot \left({x}^{2} + 2 x + 1\right) = 3 x + 6$

$\left(A + C\right) \cdot {x}^{3} + \left(- A + B + C + D\right) \cdot {x}^{2} + \left(- A - 2 B - C + 2 D\right) \cdot x + A + B - C + D = 3 x + 6$

After equating coefficients, I found

$A + C = 0$

$- A + B + C + D = 0$

$- A - 2 B - C + 2 D = 3$ and

$A + B - C + D = 6$

After solving them, I found $A = \frac{3}{2}$, $C = - \frac{3}{2}$, $D = \frac{9}{4}$ and $B = \frac{3}{4}$

Thus,

$\int \frac{3 x + 6}{\left(x + 1\right) \cdot \left(x - 1\right)} ^ 2 \cdot \mathrm{dx}$

=$\frac{3}{2} \int \frac{\mathrm{dx}}{x + 1} + \frac{3}{4} \int \frac{\mathrm{dx}}{x + 1} ^ 2 - \frac{3}{2} \int \frac{\mathrm{dx}}{x - 1} + \frac{9}{4} \int \frac{\mathrm{dx}}{x - 1} ^ 2$

=$\frac{3}{2} L n \left(x + 1\right) - \frac{3}{4 x + 4} - \frac{3}{2} L n \left(x - 1\right) - \frac{9}{4 x - 4} + C$

=$\frac{3}{2} L n \left(\frac{x + 1}{x - 1}\right) - \frac{6 x + 3}{2 {x}^{2} - 2} + C$