# How do you integrate int(x+1)/((2x-4)(x-4)(x-7)) using partial fractions?

Dec 16, 2017

$\frac{3}{20} L n \left(A b s \left(x - 2\right)\right) - \frac{5}{12} \left(A b s \left(x - 4\right)\right) + \frac{4}{15} L n \left(A b s \left(x - 7\right)\right) + C$

#### Explanation:

$\frac{x + 1}{\left(2 x - 4\right) \cdot \left(x - 4\right) \left(x - 7\right)}$

=$\frac{1}{2} \cdot \frac{x + 1}{\left(x - 2\right) \cdot \left(x - 4\right) \left(x - 7\right)}$

I decomposed $\frac{x + 1}{\left(x - 2\right) \cdot \left(x - 4\right) \left(x - 7\right)}$ into basic fractions,

$\frac{x + 1}{\left(x - 2\right) \cdot \left(x - 4\right) \left(x - 7\right)}$

=$\frac{A}{x - 2} + \frac{B}{x - 4} + \frac{C}{x - 7}$

After expanding denominator,

$A \left(x - 4\right) \left(x - 7\right) + B \left(x - 2\right) \left(x - 7\right) + C \left(x - 2\right) \left(x - 4\right) = x + 1$

Set $x = 2$, $10 A = 3$, so $A = \frac{3}{10}$

Set $x = 4$, $- 6 B = 5$, so $B = - \frac{5}{6}$

Set $x = 7$, $15 C = 8$, so $C = \frac{8}{15}$

Thus,

$\int \frac{x + 1}{\left(2 x - 4\right) \cdot \left(x - 4\right) \left(x - 7\right)} \cdot \mathrm{dx}$

=$\frac{1}{2} \int \frac{x + 1}{\left(2 x - 4\right) \cdot \left(x - 4\right) \left(x - 7\right)} \cdot \mathrm{dx}$

=$\frac{1}{2} \cdot \frac{3}{10} \int \frac{\mathrm{dx}}{x - 2} - \frac{1}{2} \cdot \frac{5}{6} \int \frac{\mathrm{dx}}{x - 4} + \frac{1}{2} \cdot \frac{8}{15} \int \frac{\mathrm{dx}}{x - 7}$

=$\frac{3}{20} \int \frac{\mathrm{dx}}{x - 2} - \frac{5}{12} \int \frac{\mathrm{dx}}{x - 4} + \frac{4}{15} \int \frac{\mathrm{dx}}{x - 7}$

=$\frac{3}{20} L n \left(A b s \left(x - 2\right)\right) - \frac{5}{12} \left(A b s \left(x - 4\right)\right) + \frac{4}{15} L n \left(A b s \left(x - 7\right)\right) + C$