# How do you evaluate the integral int (2x-3)/((x-1)(x+4))?

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#### Explanation

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#### Explanation:

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Mar 8, 2018

$\int \frac{2 x - 3}{\left(x - 1\right) \left(x + 4\right)} \mathrm{dx} = - \frac{1}{5} \ln \left\mid x - 1 \right\mid + \frac{11}{5} \ln \left\mid x + 4 \right\mid + C$

#### Explanation:

Use partial fractions decomposition:

$\frac{2 x - 3}{\left(x - 1\right) \left(x + 4\right)} = \frac{A}{x - 1} + \frac{B}{x + 4}$

$2 x - 3 = A \left(x + 4\right) + B \left(x - 1\right)$

$2 x - 3 = \left(A + B\right) x + 4 A - B$

$\left\{\begin{matrix}A + B = 2 \\ 4 A - B = - 3\end{matrix}\right.$

$\left\{\begin{matrix}5 A = - 1 \\ B = - A + 2\end{matrix}\right.$

$\left\{\begin{matrix}A = - \frac{1}{5} \\ B = \frac{11}{5}\end{matrix}\right.$

$\frac{2 x - 3}{\left(x - 1\right) \left(x + 4\right)} = - \frac{1}{5 \left(x - 1\right)} + \frac{11}{5 \left(x + 4\right)}$

$\int \frac{2 x - 3}{\left(x - 1\right) \left(x + 4\right)} \mathrm{dx} = - \frac{1}{5} \int \frac{\mathrm{dx}}{x - 1} + \frac{11}{5} \int \frac{\mathrm{dx}}{x + 4}$

$\int \frac{2 x - 3}{\left(x - 1\right) \left(x + 4\right)} \mathrm{dx} = - \frac{1}{5} \ln \left\mid x - 1 \right\mid + \frac{11}{5} \ln \left\mid x + 4 \right\mid + C$

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