Question #36feb

Apr 3, 2017

$\int \frac{2 x - 3}{x - 1} = 2 x - \ln \left\mid x - 1 \right\mid + C$

Explanation:

Write the integrand function as:

$\frac{2 x - 3}{x - 1} = 2 \left(\frac{x - \frac{3}{2}}{x - 1}\right) = 2 \left(\frac{x - 1 - \frac{1}{2}}{x - 1}\right) = 2 - \frac{1}{x - 1}$

So:

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

and using the linearity of integrals:

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