## anonymous 4 years ago find the anti-derivative

1. anonymous

$\int\limits_{}^{?}2x/(x-1)^{2}$ the ? doesn't mean anything

2. anonymous

try $u=x-1$ $du=dx$ and $x=u-1$ so $2x = 2u-2$

3. anonymous

oops sorry if $u=x-1$ then $x=u+1$ and $2x=2u+2$ my mistake

4. anonymous

good from there?

5. anonymous

yes but after that the (2x+2) *u^-2 would equal 2u^-1 +2u^-2 and ln is required for the -1 exponent and that is what i am having trouble with

6. anonymous

you should have only "u" involved. as follows $\int\frac{2u+2}{u^2}du$ then break it in to two parts

7. anonymous

$\int \frac{2u}{u^2}du+\int \frac{2}{u^2}du$ first one becomes $\int \frac{2}{u}du$ whose "anti derivative" is the log

8. anonymous

oooh i see what you are asking. you cannot use the power rule backwards when integrating $\int \frac{1}{x}dx$ that is $\frac{1}{x}dx=\ln(x)$

9. anonymous

ohhhhhh ok i understand that now thanks

10. anonymous

i meant of course $\int \frac{1}{x}dx=\ln(x)$