# What is f(x) = int x/(x-1) dx if f(2) = 0 ?

May 26, 2016

Since $\ln$ can't help you, set the denominator because of its simple form as a variable. When you solve the integral, just set $x = 2$ to fit the $f \left(2\right)$ in the equation and find the integration constant.

$f \left(x\right) = x + \ln | x - 1 | - 2$

#### Explanation:

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

The $\ln$ function will not help in this case. However, since the denominator is quite simple (1st grade):

Set $u = x - 1 \implies x = u + 1$

and $\frac{\mathrm{du}}{\mathrm{dx}} = d \frac{x + 1}{\mathrm{dx}} = \left(x + 1\right) ' = 1 \implies \frac{\mathrm{du}}{\mathrm{dx}} = 1 \iff \mathrm{du} = \mathrm{dx}$

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

$= \int \left(1 + \frac{1}{u}\right) \mathrm{du} = \int 1 \mathrm{du} + \int \frac{\mathrm{du}}{u} = u + \ln | u | + c$

Substituting $x$ back:

$u + \ln | u | + c = x - 1 + \ln | x - 1 | + c$

So:

$f \left(x\right) = \int \frac{x}{x - 1} \mathrm{dx} = x - 1 + \ln | x - 1 | + c$

$f \left(x\right) = x - 1 + \ln | x - 1 | + c$

To find $c$ we set $x = 2$

$f \left(2\right) = 2 - 1 + \ln | 2 - 1 | + c$

$0 = 1 + \ln 1 + c$

$c = - 1$

Finally:

$f \left(x\right) = x - 1 + \ln | x - 1 | + c = x - 1 + \ln | x - 1 | - 1 = x + \ln | x - 1 | - 2$

$f \left(x\right) = x + \ln | x - 1 | - 2$