Question

find the anti-derivative

Answer 1

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

Answer 2

try
\[u=x-1\]
\[du=dx\] and
\[x=u-1\] so
\[2x = 2u-2\]

Answer 3

oops sorry if
\[u=x-1\] then
\[x=u+1\] and
\[2x=2u+2\] my mistake

Answer 4

good from there?

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

ohhhhhh ok i understand that now thanks

Answer 10

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