# Thread: integral x/(x-6) using substitution

1. ## integral x/(x-6) using substitution

It's been a few months since doing any integration, need a helping hand.
Trying to solve this by substitution.

If i let u=x-6 than my integral becomes

$\displaystyle \int {xu^{ - 1} dx}$

with du = 1 dx

and so

$\displaystyle \int {xu^{ - 1} du}$

but i don't think this really helps me.

2. You got the right substitution.

Just notice that: $\displaystyle {\color{red}u = x - 6} \ \Leftrightarrow \ {\color{blue}x = u + 6}$

So: $\displaystyle \int \frac{{\color{blue}x}}{{\color{red}x-6}} \ dx = \int \frac{u + 6}{u} \ du = \int \left(1 + 6u^{-1}\right)du$

3. Thanks

4. Originally Posted by Craka
It's been a few months since doing any integration, need a helping hand.
Trying to solve this by substitution.

If i let u=x-6 than my integral becomes

$\displaystyle \int {xu^{ - 1} dx}$

with du = 1 dx

and so

$\displaystyle \int {xu^{ - 1} du}$

but i don't think this really helps me.
Alternatively, you can use long division on the integrand, eliminating the need for substitution.

Notice $\displaystyle \frac{x}{x - 6} = 1 + \frac{6}{x - 6}$.

So $\displaystyle \int{\frac{x}{x - 6}\,dx} = \int{1 + \frac{6}{x - 6}\,dx}$

$\displaystyle = x + 6\ln{(x - 6)} + C$.

5. I prefer 'short' division...

$\displaystyle {x\over x - 6} = {(x-6)+6\over x - 6}= 1+ {6\over x - 6}$.