# How do you integrate (6x^2+1)/(x^2(x-1)^2) using partial fractions?

##### 1 Answer
Oct 22, 2016

The integral is $\int \frac{\left(6 {x}^{2} + 1\right) \mathrm{dx}}{{x}^{2} {\left(x - 1\right)}^{2}} = - \frac{1}{x} + 2 \ln x - 2 \ln \left(x - 1\right) - \frac{7}{x - 1} + C$

#### Explanation:

Let's first determine the coefficients of the partial fractions

$\frac{6 {x}^{2} + 1}{{x}^{2} {\left(x - 1\right)}^{2}} = \frac{A}{x} ^ 2 + \frac{B}{x} + \frac{C}{x - 1} + \frac{D}{x - 1} ^ 2$

$\frac{6 {x}^{2} + 1}{{x}^{2} {\left(x - 1\right)}^{2}} = \frac{A {\left(x - 1\right)}^{2} + B x {\left(x - 1\right)}^{2} + C {x}^{2} \left(x - 1\right) + D {x}^{2}}{{x}^{2} {\left(x - 1\right)}^{2}}$

$\left(6 {x}^{2} + 1\right) = A {\left(x - 1\right)}^{2} + B x {\left(x - 1\right)}^{2} + C {x}^{2} \left(x - 1\right) + D {x}^{2}$

Let x=0, then $1 = A + 0 + 0 + 0$ ; So $A = 1$
Let x=1, then $7 = D$ ; so $D = 7$
Also,coefficients of ${x}^{3}$; $0 = B + C$, ; so $B = - C$
And coefficients of ${x}^{2}$ ; $6 = A - 2 B - C + D$
So $6 = 1 + 2 C - C + 7$ $\implies$$C = - 2$
And $B = 2$

$\frac{6 {x}^{2} + 1}{{x}^{2} {\left(x - 1\right)}^{2}} = \frac{1}{x} ^ 2 + \frac{2}{x} - \frac{2}{x - 1} + \frac{7}{x - 1} ^ 2$

So

int((6x^2+1)dx)/(x^2(x-1)^2)=int(1dx)/x^2+int(2dx)/x-int(2dx)/(x-1)+ (7dx)/(x-1)^2

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