# How do you integrate int(x+1)/((6x^2+4)(x-5)) using partial fractions?

Oct 2, 2017

See below.

#### Explanation:

1) I used basic fractions method.

Inıtıially, I decomposed integrand into basic fractions,

$\frac{x + 1}{\left(6 {x}^{2} + 4\right) \cdot \left(x - 5\right)} = \frac{A}{x - 5} + \frac{B x + C}{6 {x}^{2} + 4}$

After expanding denominator,

$A \cdot \left(6 {x}^{2} + 4\right) + \left(B x + C\right) \cdot \left(x - 5\right) = x + 1$

$\left(6 A + B\right) \cdot {x}^{2} + \left(C - 5 B\right) \cdot x + 4 A - 5 C = x + 1$

After equating coefficients, I found

$6 A + B = 0$, $C - 5 B = 1$ and $4 A - 5 C = 1$ equations.

After solving them, $A = \frac{3}{77} , B = - \frac{18}{77} \mathmr{and} C = - \frac{13}{77}$

Thus,

$\int \frac{\left(x + 1\right) \mathrm{dx}}{\left(6 {x}^{2} + 4\right) \cdot \left(x - 5\right)}$

=$\frac{3}{77}$$\int \frac{\mathrm{dx}}{x - 5}$-$\frac{1}{77}$$\int \frac{\left(18 x + 13\right) \cdot \mathrm{dx}}{6 {x}^{2} + 4}$

=$\frac{3}{77} \cdot \ln \left(x - 5\right)$-$\frac{3}{154}$$\int \frac{12 x \cdot \mathrm{dx}}{6 {x}^{2} + 4}$-$\frac{13}{77}$$\int \frac{\mathrm{dx}}{6 {x}^{2} + 4}$

=$\frac{3}{77} \cdot \ln \left(x - 5\right) - \frac{3}{154} \cdot \ln \left(6 {x}^{2} + 4\right) - \frac{13 \sqrt{6}}{924} \cdot \arctan \left(\frac{3 x}{\sqrt{6}}\right) + C$