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

Dec 19, 2017

The integral equals $2 x + 2 \ln | x - 1 | + C$

#### Explanation:

I would use partial fractions.

$\frac{A}{1} + \frac{B}{x - 1} = \frac{2 x}{x - 1}$

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

$A x - A + B = 2 x$

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

From here it's clear the $A = 2$ and $B - A = 0$ therefore, $B = 2$.

$I = \int 2 + \frac{2}{x - 1} \mathrm{dx}$

$I = 2 x + 2 \ln | x - 1 | + C$

Hopefully this helps!

Dec 19, 2017

Another way to see the rewrite.

#### Explanation:

$\int \frac{2 x \mathrm{dx}}{x - 1} \mathrm{dx} = 2 \int \frac{x}{x - 1} \mathrm{dx}$

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

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

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

$= 2 \left(x + \ln \left\mid x - 1 \right\mid\right) + C$