Question

Integrating problem x/(1-x)^2

Answer 1

Wait me show my steps

Answer 2

u sub right ?

Answer 3

\[x=\tan \theta\]\[dx=\sec ^{2}\theta d \theta\]
\[\int\limits_{}^{}\frac{\tan \theta}{(1-\tan \theta)^{2}}\sec ^{2}\theta d \theta\]
\[\int\limits\limits\limits_{}^{}\frac{\tan \theta \sec ^{2}\theta}{(1+\tan ^{2} \theta-2\tan \theta)}\ d \theta\]
\[\int\limits\limits\limits\limits_{}^{}\frac{\tan \theta \sec ^{2}\theta}{(\sec ^{2}\theta-2\tan \theta)}\ d \theta\]
Im stucked here

Answer 4

try
\[u=1-x\] first

Answer 5

have you learned partial fractions yet? thats the method I used on this.

Answer 6

with u=1-x ;)

Answer 7

er x-1 rather.

Answer 8

\[u=1-x, x=1-u\]
\[-du=dx\]

Answer 9

you get
\[-\int \frac{1-u}{u^2}du\]

Answer 10

you can also do it by partial fractions, and also by a gimmick

Answer 11

Oh okay, i get the answer, but the problem is with myself ,
i dont know when to use the trigonometry :(

Answer 12

\[\int \frac{x}{(x-1)^2} dx = \int (\frac{1}{x-1}+\frac1{(x-1)^2})dx\]
let:
u = x-1
du=dx
plug and chug.

Answer 13

@green that would be probably the most work

Answer 14

probably, but i like partial fractions :)

Answer 15

more than one way to skin a cat

Answer 16

plus you skipped a step !

Answer 17

i dont know when to use the trigonometry :(

Answer 18

@Calculator Once you know all the methods of integration, it takes some time to get to know when to use what--and even then, people argue about when to do what... kinda like me and @satellite73 ... fairly often, lol

Answer 19

how do you know
\[\frac{x}{(1-x)^2}=\frac{1}{1-x}+\frac{1}{(1-x)^2}\] and is it even true??

Answer 20

partial
fractions
definition
QED

Answer 21

to be honest you usually use a trig sub when you come to the section in the book on trig subs

Answer 22

but the thing is in the exam they didnt tell you use which technique, lol

Answer 23

then you will see some similar forms after that

Answer 24

True, but at a certain point, you need to use all of the methods... if you have a worthwhile calc II teacher they will construct problems that utilize all the methods in a single question to truly test you.

Answer 25

yeah nothing more annoying that techniques of integration
bunch of teachers showing off
look we can do this techninque
look we can do that one
truth is almost all functions are not integrable

Answer 26

lol i dont even use integration anymore, functions that are continuous and easily defined are so rare in actual applications.

Answer 27

i don't know if it works, you can use trigo sub when you see the form \[\sqrt{a ^{2}-x ^{2}}\]or\[\sqrt{x ^{2}-a ^{2}}\] or \[\sqrt{a ^{2}+x ^{2}}\] (ps i'm really bad in dealing with integration)