# Thread: Integration by Substitution question?

1. ## Integration by Substitution question?

Can you please show me the steps to integrate:

∫(x/2-x) dx ?? i let u = (2-x) and then??

Help is very appreciated! thanks!!

2. Originally Posted by Yehia
Can you please show me the steps to integrate:

∫(x/2-x) dx ?? i let u = (2-x) and then??
Then differentiate $u=2-x$ to get $du=-dx,$ besides $x=2-u.$ Can you take it?

3. Originally Posted by Krizalid
Then differentiate $u=2-x$ to get $du=-dx,$ besides $x=2-u.$ Can you take it?
I do do that! to get: ∫ (u-2)/u du

and that simplifies to : ∫ 1 - 2u^-1 du

still stumped though :S

4. Well done, you almost got it. (Oh, the numerador is actually $2-u.$)

Next step is to note that $\int du=u+k$ and $\int\frac{du}u=\ln|u|+m.$

5. Originally Posted by Krizalid
Well done, you almost got it. (Oh, the numerador is actually $2-u.$)

Next step is to note that $\int du=u+k$ and $\int\frac{du}u=\ln|u|+m.$

But the numerator is MINUS x isnt it? and if x = 2 - u, then -x = u - 2 ? if im not wrong?

And i dont understand your second bit (which is right btw, according to the book) can you explain?

thanks a lot

6. Ah yes, you're right, I forgot the minus sign when differentiating.