# How do you integrate int (4x+3) / (x - 1)^2dx using partial fractions?

May 22, 2017

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#### Explanation:

$\int \frac{4 \left(x - 1\right) + 4 + 3}{x - 1} ^ 2 \mathrm{dx}$

By substituting $x$ in $4 x$ with the denominator, the fraction is split into partial fractions. Include a $+ 4$ so the original fraction isn't changed.

$\int \frac{4 \cancel{x - 1}}{x - 1} ^ \cancel{2} + \frac{7}{x - 1} ^ 2 \mathrm{dx}$

Split the integrals to make it easier.

$= \int \frac{4}{x - 1} \mathrm{dx} + \int \frac{7}{x - 1} ^ 2 \mathrm{dx}$

$= 4 \int \frac{1}{x - 1} \mathrm{dx} + 7 \int {\left(x - 1\right)}^{-} 2 \mathrm{dx}$

1. $4 \int \frac{1}{x - 1} \mathrm{dx} = 4 \ln | x - 1 | + {c}_{1}$ Integration of inverse functions.

2. $7 \int {\left(x - 1\right)}^{-} 2 \mathrm{dx} = 7 \cdot - 1 \cdot {\left(x - 1\right)}^{-} 1 + {c}_{2}$ Using reverse chain rule.

Therefore,

$= 4 \int \frac{1}{x - 1} \mathrm{dx} + 7 \int {\left(x - 1\right)}^{-} 2 \mathrm{dx}$

$= 4 \ln | x - 1 | - 7 {\left(x - 1\right)}^{-} 1 + {c}_{3}$