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

Jan 19, 2017

The answer is =-1/x+2ln(∣x∣)-ln(∣x+3∣)+C

#### Explanation:

Let's perform the decomposition into partial fractions

$\frac{{x}^{2} + 7 x + 3}{{x}^{2} \left(x + 3\right)} = \frac{A}{x} ^ 2 + \frac{B}{x} + \frac{C}{x + 3}$

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

Equalising the denominators

${x}^{2} + 7 x + 3 = A \left(x + 3\right) + B x \left(x + 3\right) + C {x}^{2}$

Let $x = 0$, $\implies$, $3 = 3 A$, $\implies$, $A = 1$

Let $x = - 3$, $\implies$, $- 9 = 9 C$, $\implies$, $C = - 1$

Coefficients of ${x}^{2}$

$1 = B + C$, $\implies$, $B = 1 - C = 2$

Therefore,

$\frac{{x}^{2} + 7 x + 3}{{x}^{2} \left(x + 3\right)} = \frac{1}{x} ^ 2 + \frac{2}{x} - \frac{1}{x + 3}$

So,

$\int \frac{\left({x}^{2} + 7 x + 3\right) \mathrm{dx}}{{x}^{2} \left(x + 3\right)} = \int \frac{\mathrm{dx}}{x} ^ 2 + 2 \int \frac{\mathrm{dx}}{x} - \int \frac{\mathrm{dx}}{x + 3}$

=-1/x+2ln(∣x∣)-ln(∣x+3∣)+C