# How do you integrate (10x + 2)/ (x^3 - 5x^2 + x - 5) using partial fractions?

Jun 25, 2016

$\int \frac{10 x + 2}{{x}^{3} - 5 {x}^{2} + x - 5} \mathrm{dx} = 2 \ln | x - 5 | - \ln \left({x}^{2} + 1\right) + K .$

#### Explanation:

We first factorise the $D r .$

$D r . = {x}^{3} - 5 {x}^{2} + x - 5 = {x}^{2} \left(x - 5\right) + 1 \left(x - 5\right) = \left(x - 5\right) \left({x}^{2} + 1\right) .$

Thus, $D r .$ has a linear factor = (x-5), and a quadratic one = (x^2+1), which can not be further factorised [ in $\mathbb{R}$]. In such a case, we need to find $A , B , C \in \mathbb{R}$ under the following supposition ::

$\frac{10 x + 2}{{x}^{3} - 5 {x}^{2} + x - 5} = \frac{10 x + 2}{\left(x - 5\right) \left({x}^{2} + 1\right)} = \frac{A}{x - 5} + \frac{2 B x + C}{{x}^{2} + 1} \ldots \ldots \ldots \ldots \ldots . . \left(1\right)$

The $2$ in $\left(2 B x + C\right)$ has been so chosen, bcz. $\frac{d}{\mathrm{dx}} \left({x}^{2} + 1\right) = 2 x .$

Simplifying the $R . H . S .$ of $\left(1\right)$ and then comparing the $N r s .$ of both sides, we find that,

$A \left({x}^{2} + 1\right) + \left(2 B x + C\right) \left(x - 5\right) = 10 x + 2. \ldots \ldots \ldots \ldots . . \left(2\right)$

To determine $A , B , C$ of $\left(2\right)$, we can take any $t h r e e$ values of x; sub. them one by one in $\left(2\right)$ giving $3$ eqns. in $A , B , C$ and solve them.

But, to make computational work easier, we tactfully choose $x = 5 , 0 , 1 ,$ to see :

$x = 5 \Rightarrow 26 A = 52 \Rightarrow A = 2.$
$x = 0 \Rightarrow A - 5 C = 2 \Rightarrow 2 - 5 C = 2 \Rightarrow C = 0.$
$x = 1 \Rightarrow 2 A + \left(2 B + C\right) \left(- 4\right) = 12 \Rightarrow 4 - 8 B = 12 \Rightarrow B = - 1.$

Using these $A , B , C$ in $\left(1\right) ,$ we have,

$\int \frac{10 x + 2}{{x}^{3} - 5 {x}^{2} + x - 5} \mathrm{dx} = \int \left[\frac{2}{x - 5} - \frac{2 x}{{x}^{2} + 1}\right] \mathrm{dx} = \int \frac{2}{x - 5} \mathrm{dx} - \int \frac{\frac{d}{\mathrm{dx}} \left({x}^{2} + 1\right)}{{x}^{2} + 1} \mathrm{dx} = 2 \ln | x - 5 | - \ln \left({x}^{2} + 1\right) + K .$
[$K$ const. of int.]

In the second integral, we have made use of the formula : $\int \frac{f ' \left(x\right)}{f} \left(x\right) \mathrm{dx} = \ln | f \left(x\right) | + {C}_{1.}$[$C - 1$ const. of int.]