# How do you find int (5x^2+3x-2)/(x^3+2x^2) dx using partial fractions?

Dec 11, 2016

$2 \ln x + 3 \ln \left(x + 2\right) + \frac{1}{x} + C$

#### Explanation:

Assume that you can write the fraction as $\frac{A x + B}{x} ^ 2 + \frac{C}{x + 2}$.
That is, $5 {x}^{2} + 3 x - 2 \equiv \left(A x + B\right) \left(x + 2\right) + C {x}^{2}$

Then, equating coefficents (or however you prefer)
you get $5 = A + C$, $3 = 2 A + B$ and $- 2 = 2 B$. Hence $B = - 1$, $A = 2$ and $C = 3$
So now you have $\int \frac{2 x - 1}{x} ^ 2 + \frac{3}{x} + 2 \mathrm{dx}$
=$\int \frac{2}{x} - \frac{1}{x} ^ 2 + \frac{3}{x + 2} \mathrm{dx}$ giving the answer above.