# How do you use partial fractions to find the integral int (x^3)/(x^2-4)^2dx?

Mar 31, 2018

The answer is $= \frac{1}{2} \ln \left(| {x}^{2} - 4 |\right) - \frac{2}{{x}^{2} - 4} + C$

#### Explanation:

Perform the decomposition into partial fractions

${x}^{3} / {\left({x}^{2} - 4\right)}^{2} = \frac{{x}^{3}}{{\left(x - 2\right)}^{2} {\left(x + 2\right)}^{2}}$

$= \frac{A}{x - 2} ^ 2 + \frac{B}{x - 2} + \frac{C}{x + 2} ^ 2 + \frac{D}{x + 2}$

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

The denominators are the same, compare the numerators

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

Let $x = 2$, $\implies$, $8 = 16 A$, $\implies$, $A = \frac{1}{2}$

Let $x = - 2$, $\implies$, $- 8 = 16 C$, $\implies$, $C = - \frac{1}{2}$

Coefficients of ${x}^{3}$

$1 = B + D$

Coefficients of ${x}^{2}$

$0 = A + 2 B + C - 2 D$

$B - D = 0$

$B = B = \frac{1}{2}$

${x}^{3} / {\left({x}^{2} - 4\right)}^{2} = \frac{\frac{1}{2}}{x - 2} ^ 2 + \frac{\frac{1}{2}}{x - 2} + \frac{- \frac{1}{2}}{x + 2} ^ 2 + \frac{\frac{1}{2}}{x + 2}$

Therefore, the integral is

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

$= - \frac{1}{2} \cdot \frac{1}{x - 2} + \frac{1}{2} \ln \left(| x - 2 |\right) + \frac{1}{2} \cdot \frac{1}{x + 2} + \frac{1}{2} \ln \left(| x + 2 |\right) + C$

$= \frac{1}{2} \ln \left(| {x}^{2} - 4 |\right) - \frac{2}{{x}^{2} - 4} + C$

Mar 31, 2018

$\frac{1}{2} \ln | \left({x}^{2} - 4\right) | - \frac{2}{{x}^{2} - 4} + C$.

#### Explanation:

As a Second Method, let us solve the Problem without

applying the Method of Partial Fraction.

Consider the subst. ${x}^{2} = y , \text{ so that, } 2 x \mathrm{dx} = \mathrm{dy}$.

$\therefore I = \int {x}^{3} / {\left({x}^{2} - 4\right)}^{2} \mathrm{dx} = \frac{1}{2} \int \frac{{x}^{2} \cdot 2 x}{{x}^{2} - 4} ^ 2 \mathrm{dx}$,

$= \frac{1}{2} \int \frac{y}{y - 4} ^ 2 \mathrm{dy}$,

$= \frac{1}{2} \int \frac{\left(y - 4\right) + 4}{y - 4} ^ 2 \mathrm{dy}$,

$= \frac{1}{2} \int \left\{\frac{y - 4}{y - 4} ^ 2 + \frac{4}{y - 4} ^ 2\right\} \mathrm{dy}$,

$= \frac{1}{2} \left\{\int \frac{1}{y - 4} \mathrm{dy} + 4 \int \frac{1}{y - 4} ^ 2 \mathrm{dy}\right\}$,

$= \frac{1}{2} \left\{\ln | \left(y - 4\right) | + 4 \cdot {\left(y - 4\right)}^{- 2 + 1} / \left(- 2 + 1\right)\right\}$.

$= \frac{1}{2} \ln | \left(y - 4\right) | - \frac{2}{y - 4}$.

Returning to ${x}^{2} = y$, we get,

$I = \frac{1}{2} \ln | \left({x}^{2} - 4\right) | - \frac{2}{{x}^{2} - 4} + C$, as Respected Narad T. has