# How do you integrate 6/(x^2-25)^2 using partial fractions?

Jun 26, 2018

$\frac{3}{250} \left[\ln | \frac{x + 5}{x - 5} | - \frac{10}{{x}^{2} - 25}\right] + C$

#### Explanation:

$\frac{6}{{x}^{2} - 25} ^ 2 = \frac{6}{{\left(x - 5\right)}^{2} {\left(x + 5\right)}^{2}}$

You can carry out the partial fractions decomposition by following the standard method, i.e., by starting with the form

$\frac{6}{{\left(x - 5\right)}^{2} {\left(x + 5\right)}^{2}} = \frac{A}{x - 5} + \frac{B}{x + 5}$
$q \quad q \quad q \quad q \quad q \quad q \quad q \quad q \quad + \frac{C}{x - 5} ^ 2 + \frac{D}{x + 5} ^ 2$

and rewriting this as

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

$q \quad q \quad + C {\left(x + 5\right)}^{2} + D {\left(x - 5\right)}^{2}$

• substituting $x = + 5$ gives $100 C = 6 \implies C = \frac{3}{50}$
• substituting $x = - 5$ gives $100 D = 6 \implies D = \frac{3}{50}$
• comparing coefficients of ${x}^{3}$ on both sides yield $A + B = 0$
• comparing coefficients of ${x}^{0}$ on both sides yield
$- {5}^{3} A + {5}^{3} B + {5}^{2} C + {\left(- 5\right)}^{2} D = 6 \implies$
$A - B = \frac{C + D}{5} - \frac{6}{5} ^ 3 = - \frac{3}{125} \implies$
$A = - B = - \frac{3}{250}$

Thus

$\frac{6}{{x}^{2} - 25} ^ 2 = - \frac{3}{250 \left(x - 5\right)} + \frac{3}{250 \left(x + 5\right)}$
$q \quad q \quad q \quad q \quad q \quad q \quad q \quad q \quad + \frac{3}{50 {\left(x - 5\right)}^{2}} + \frac{3}{50 {\left(x + 5\right)}^{2}}$

Thus the required integral is

$\int \frac{6 \mathrm{dx}}{{x}^{2} - 25} ^ 2 = - \int \frac{3 \mathrm{dx}}{250 \left(x - 5\right)} + \int \frac{3 \mathrm{dx}}{250 \left(x + 5\right)}$
$q \quad q \quad q \quad q \quad q \quad q \quad q \quad q \quad + \int \frac{3 \mathrm{dx}}{50 {\left(x - 5\right)}^{2}} + \int \frac{3 \mathrm{dx}}{50 {\left(x + 5\right)}^{2}}$
$q \quad q \quad = \frac{3}{250} \ln | \frac{x + 5}{x - 5} | - \frac{3}{50} \left[\frac{1}{x - 5} + \frac{1}{x + 5}\right] + C$
$q \quad q \quad = \frac{3}{250} \left[\ln | \frac{x + 5}{x - 5} | - \frac{10}{{x}^{2} - 25}\right] + C$