# How do you integrate (5x^2+7x-4)/(x^3+4x^2) using partial fractions?

Nov 4, 2016

The answer is $= \frac{1}{x} + 2 \ln x + 3 \ln \left(x + 4\right) + C$

#### Explanation:

let's simplify the denominator ${x}^{3} + 4 {x}^{2} = {x}^{2} \left(x + 4\right)$
So $\frac{{x}^{2} + 7 x - 4}{{x}^{2} \left(x + 4\right)} = \frac{A}{x} ^ 2 + \frac{B}{x} + \frac{C}{x + 4}$

$\therefore 5 {x}^{2} + 7 x - 4 = A \left(x + 4\right) + B x \left(x + 4\right) + C {x}^{2}$
$- 4 = 4 A$$\implies$ $A = - 1$
coefficients of x, $7 = A + 4 B$$\implies$$B = 2$
Coefficients of x^2, $5 = B + C$$\implies$ $C = 3$
$\therefore \int \frac{\left({x}^{2} + 7 x - 4\right) \mathrm{dx}}{{x}^{2} \left(x + 4\right)} = \int - \frac{\mathrm{dx}}{x} ^ 2 + \int \frac{2 \mathrm{dx}}{x} + \int \frac{3 \mathrm{dx}}{x + 4}$
$= \frac{1}{x} + 2 \ln x + 3 \ln \left(x + 4\right) + C$