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

Nov 25, 2015

1) make sure that the degree of your denominator is greater than the degree of your numerator
2) factorize the denominator
3) perform the partial fraction decomposition
4) solve the integral!

Solution: $x + 8 \ln | x - 2 | - 5 \ln | x - 1 | + c$

#### Explanation:

1) Check degrees of denominator and numerator

First of all, you can't immediately start building partial fractions since the degree of your numerator is equal to the degree of your denominator (the strongest power in both expressions is ${x}^{2}$).

However, the partial fraction decomposition will only work if the degree of the denominator is greater than the degree of the numerator.

If the expression was more complicated, normally, at this point, a polynomial division is in order. Here we can achieve the goal in an easier way:

${x}^{2} / \left({x}^{2} - 3 x + 2\right) = \frac{{x}^{2} \textcolor{b l u e}{- 3 x + 2} \textcolor{red}{+ 3 x - 2}}{{x}^{2} - 3 x + 2}$

$\textcolor{w h i t e}{\times \times \times \times x} = \frac{{x}^{2} - 3 x + 2}{{x}^{2} - 3 x + 2} + \frac{3 x - 2}{{x}^{2} - 3 x + 2}$

$\textcolor{w h i t e}{\times \times \times \times x} = 1 + \frac{3 x + 2}{{x}^{2} - 3 x + 2}$

So, now we can build the partial fractions decomposition of the latter fraction.

2) Complete factorization of the denominator

To do so, let's factorize the denominator completely first:

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

[You can do so by setting ${x}^{2} - 3 x + 2 = 0$, finding the solutions ${x}_{1}$ and ${x}_{2}$ and factorizing with $\left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$.]

3) Partial fraction decomposition

So, our goal is to find $A$ and $B$ so that

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

First, multiply both sides of the equation with $\left(x - 2\right) \left(x - 1\right)$.

$\iff 3 x + 2 = A \cdot \left(x - 1\right) + B \cdot \left(x - 2\right)$
$\iff \textcolor{b l u e}{3 x} + \textcolor{red}{2} = \textcolor{b l u e}{A \cdot x} \textcolor{red}{\text{ "- A) + color(blue)(B * x) color(red)(" } - 2 B}$

To find the solution of this equation, we need to "collect" the $\textcolor{b l u e}{x}$ terms and the $\textcolor{red}{\text{constant}}$ terms:

$\left\{\begin{matrix}3 = A + B \textcolor{w h i t e}{\times \times \times x} \textcolor{b l u e}{x} \text{ terms" \\ 2 = -A - 2B color(white)(xxxx) color(red)("constant")" terms}\end{matrix}\right.$

The solution of this linear equation is $B = - 5$ and $A = 8$.

So, our partial fraction decomposition is:

$\frac{3 x + 2}{\left(x - 2\right) \left(x - 1\right)} = \frac{8}{x - 2} - \frac{5}{x - 1}$

In total:

${x}^{2} / \left({x}^{2} - 3 x + 2\right) = 1 + \frac{3 x + 2}{{x}^{2} - 3 x + 2} = 1 + \frac{8}{x - 2} - \frac{5}{x - 1}$

4) Solving the integral

The last thing left to do is solve the integral!

$\int {x}^{2} / \left({x}^{2} - 3 x + 2\right) \text{d" x = int (1 + 8 / (x-2) - 5 / (x-1)) "d} x$

$\textcolor{w h i t e}{\times \times \times \times \times \times} = \int 1 \text{d"x + int 8/(x-2) "d" x - int 5/(x-1) "d} x$

$\textcolor{w h i t e}{\times \times \times \times \times \times} = x + 8 \ln | x - 2 | - 5 \ln | x - 1 | + c$

Hope that this helped!