# How do you use partial fractions to find the integral int (6x^2+1)/(x^2(x-1)^3)dx?

Oct 12, 2017

$- 3 \ln x + \frac{1}{x} - 3 \ln \left(x - 1\right) + \frac{2}{x - 1} - \frac{7}{2} \cdot {\left(x - 1\right)}^{- 2} + C$

#### Explanation:

I decomposed integrand into basic fractions,

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

=$- \frac{3}{x} - \frac{1}{x} ^ 2 + \frac{3}{x - 1} - \frac{2}{x - 1} ^ 2 + \frac{7}{x - 1} ^ 3$

Hence,

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

=$- \int \frac{3 \mathrm{dx}}{x}$-$\int \frac{\mathrm{dx}}{x} ^ 2$-$\int \frac{3 \mathrm{dx}}{x - 1}$-$\int \frac{2 \mathrm{dx}}{x - 1} ^ 2$+$\int \frac{7 \mathrm{dx}}{x - 1} ^ 3$

=$- 3 \ln x + \frac{1}{x} - 3 \ln \left(x - 1\right) + \frac{2}{x - 1} - \frac{7}{2} \cdot {\left(x - 1\right)}^{- 2} + C$