# How do you integrate (x^2 + 33)/(x^3 + x^2) using partial fractions?

Nov 13, 2016

$y = 34 \ln | x + 1 | - 33 \ln | x | - \frac{33}{x} + C$

#### Explanation:

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

So,

$\frac{A x + B}{x} ^ 2 + \frac{C}{x + 1} = \frac{{x}^{2} + 33}{\left({x}^{2}\right) \left(x + 1\right)}$

$\left(A x + B\right) \left(x + 1\right) + C {x}^{2} = {x}^{2} + 33$

$A {x}^{2} + B x + A x + B + C {x}^{2} = {x}^{2} + 33$

$\left(A + C\right) {x}^{2} + \left(A + B\right) x + B = {x}^{2} + 33$

So,

$\left\{\begin{matrix}A + C = 1 \\ A + B = 0 \\ B = 33\end{matrix}\right.$

Solving, we get that $A = - 33 , B = 33 , C = 34$.

Hence, the partial fraction decomposition is as follows:

$\frac{- 33 x + 33}{x} ^ 2 + \frac{34}{x + 1}$

We can integrate the second term as $34 \ln | x + 1 | + C$. However, we can decompose the first term further.

$\int \frac{33 - 33 x}{x} ^ 2 \mathrm{dx} = \int \frac{33 \left(1 - x\right)}{x} ^ 2 \mathrm{dx} = 33 \int \frac{1 - x}{x} ^ 2 \mathrm{dx}$

$\frac{A}{x} + \frac{B}{x} ^ 2 = \frac{1 - x}{x} ^ 2$

$A x + B = 1 - x$

Here, we have that $A = - 1$ and $B = 1$.

Hence,

$\implies 33 \int \left(\frac{1}{x} ^ 2 - \frac{1}{x}\right) \mathrm{dx}$

$\implies 33 \int \left({x}^{- 2} - \frac{1}{x}\right) \mathrm{dx}$

$\implies 33 \left(- \frac{1}{x} - \ln | x |\right) + C$

Putting this and the part that we integrated above together gives

$\implies - \frac{33}{x} - 33 \ln | x | + 34 \ln | x + 1 | + C$

Hopefully this helps!  