# How do you integrate (6x^2 + 1) /( (x^2)(x-1)^3) using partial fractions?

$\frac{1}{x} + \frac{2}{x - 1} - \frac{14}{x - 1} ^ 2 + 3 \setminus \ln | \frac{x - 1}{x} | + C$

#### Explanation:

Let

$\setminus \frac{6 {x}^{2} + 1}{{x}^{2} {\left(x - 1\right)}^{3}} = \frac{A}{x - 1} + \frac{B}{x - 1} ^ 2 + \frac{C}{x - 1} ^ 3 + \frac{D}{x} + \frac{E}{{x}^{2}}$

$\setminus \frac{6 {x}^{2} + 1}{{x}^{2} {\left(x - 1\right)}^{3}} = \frac{A {x}^{2} {\left(x - 1\right)}^{2} + B {x}^{2} \left(x - 1\right) + C {x}^{2} + \left(D x + E\right) {\left(x - 1\right)}^{3}}{{x}^{2} {\left(x - 1\right)}^{3}}$

$6 {x}^{2} + 1 = \left(A + D\right) {x}^{4} + \left(- 2 A + B - 3 D + E\right) {x}^{3} + \left(A - B + C + 3 D - 3 E\right) {x}^{2} + \left(- D + 3 E\right) x - E$

Comparing the corresponding coefficients on both the sides we get

$A + D = 0 \setminus \ldots \ldots \ldots \left(1\right)$

$- 2 A + B - 3 D + E = 0 \setminus \ldots \ldots \ldots \left(2\right)$

$A - B + C + 3 D - 3 E = 6 \setminus \ldots \ldots \ldots \ldots \left(3\right)$

$- D + 3 E = 0 \setminus \ldots \ldots \ldots \ldots \left(4\right)$

$E = - 1 \setminus \ldots \ldots \ldots . \left(5\right)$

Solving all above five linear equations, we get

$A = 3 , B = - 2 , C = 7 , D = - 3 , E = - 1$

Now, setting above values , the partial fractions are given as follows

$\setminus \frac{6 {x}^{2} + 1}{{x}^{2} {\left(x - 1\right)}^{3}} = \frac{3}{x - 1} - \frac{2}{x - 1} ^ 2 + \frac{7}{x - 1} ^ 3 - \frac{3}{x} - \frac{1}{x} ^ 2$

Now, integrating above equation w.r.t. $x$, we get

$\setminus \int \setminus \frac{6 {x}^{2} + 1}{{x}^{2} {\left(x - 1\right)}^{3}} \setminus \mathrm{dx} = \setminus \int \frac{3}{x - 1} \setminus \mathrm{dx} - \setminus \int \frac{2}{x - 1} ^ 2 \setminus \mathrm{dx} + \setminus \int \frac{7}{x - 1} ^ 3 \setminus \mathrm{dx} - \setminus \int \frac{3}{x} \setminus \mathrm{dx} - \setminus \int \frac{1}{x} ^ 2 \setminus \mathrm{dx}$

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

$= 3 \setminus \ln | x - 1 | + \frac{2}{x - 1} - \frac{14}{x - 1} ^ 2 - 3 \setminus \ln | x | + \frac{1}{x} + C$

$= \frac{1}{x} + \frac{2}{x - 1} - \frac{14}{x - 1} ^ 2 + 3 \setminus \ln | \frac{x - 1}{x} | + C$