Question

Integral question (see inside)

Answer 1

\[
\int\sqrt{\frac{x-1}{x+1}}\cdot\frac{1}{x^2}dx
\]

Answer 2

-1 + 2x - 2x^2 + ..... \(-1+2\sum_{n=1}^{inf}x^n\ \) ; :)
-------------
1+x ) -1+x
(-1-x)
-------
2x
(2x+2x^2)
-----------
-2x^2

Answer 3

Didn't really follow what you were doing there. For context, this problem is intended to be done with u-substitution, but is starred as being difficult.

Answer 4

im just looking at it from a different perspective to see if it makes it simpler; it doesnt so far, but its good to stretch the grey matter some.

Answer 5

\[\int \frac{1}{x^2}\sqrt{-1+2\sum_1(-1)^{n-1}x^n\ }\ dx\]
\[\int \sqrt{-x^{-4}+2\sum_1(-1)^{n-1}x^{n-4}\ }\ dx\]

lol, take u = the under stuff maybe? :)

Answer 6

Haha I don't even know what to begin to do with that :D

I tried u=\(\sqrt{x+1}\), and I tried conjugating then doing \(u=\sqrt{x^2-1}\). The second one was totally a bust, the first one might have helped but I couldn't really tell haha.

Answer 7

gotta love the "aw snap" page in the middle of a complicated write up :/

Answer 8

substitute \(u=\frac{1}{x}\) and then \(v=u+1\)

Answer 9

try rewriteing that fractional part in a few different ways

x+1/x-1 = 2x/(x+1) - 1 for instance

Answer 10

With @ljiensen's strategy I get to \[\int\sqrt{\frac{2-v}{v}}\], if I did it right.

Answer 11

\[u=\frac{2x}{x+1}-1\]
\[\frac{du}{dx}=\frac{2}{(x+1)^2}\]
\[dx=\frac{(x+1)^2}{2}du\]
\[\sqrt{u}*\frac{(x+1)^2}{2x^2}\]
\[\sqrt{u}*(\frac{x+1}{2x})^2\]
\[\sqrt{u}*(\frac{1}{2}+\frac{1}{2x})^2\]

\[\frac{u+1}{1-u}=x\]

\[\sqrt{u}*(\frac{1}{2}+\frac{1-u}{2(u+1)})^2\]
\[\sqrt{u}*(\frac{1}{u+1})^2\]
\[(\frac{u}{u+1})^2\]

maybe ... :)

Answer 12

\[\int u^2(u+1)^{-2}du=-u^2(u+1)^{-1}-2\int u(u+1)^{-1}du\]\[\hspace{3em}=-u^2(u+1)^{-1}-2(\ uln(u+1)-\int ln(u+1)du)\]

Answer 13

use
1/x=t

Answer 14

nono

Answer 15

@vamgadu go vam

Answer 16

multiply numerator and denominator with sqrt(1-t)

Answer 17

it makes life easier try it @nbouscal.

Answer 18

So we'll still have\[
\int\frac{\cos t\sqrt{1-t}}{(1-t)\sqrt{1+\sin^2t}}dt
\]

Answer 19

no dont use sint

Answer 20

Oh, okay. Yeah, I was already using 1/x from @ljensen suggestion earlier. The sin^2(t) part was what I was saying made it more complex lol

Answer 21

With just the 1/x substitution we have:\[
\int\sqrt{\frac{1-u}{1+u}}du
\]

Answer 22

we have
\[\sqrt{1-t/1+t}\]
multiply it with \[\sqrt{1-t}\]
it becomes
\[(1/\sqrt{1-t^{2}})-(t/\sqrt{1-t^{2}})\]

Answer 23

Okay, yes, give me a moment I think I can work this out.

Answer 24

\[\sin^{-1} t +.5 \sqrt{1-t^{2}}\]

Answer 25

So after substitutions we are left with \(\arcsin\frac{1}{x}+\sqrt{1-(\frac{1}{x})^2}\)

Answer 26

Had lost a negative sign, actually \(-\arcsin\frac{1}{x}-\sqrt{1-(\frac{1}{x})^2}\), I think.

Answer 27

WolframAlpha comes up with something much different, but it could be a different way of writing it. Always hard to tell. I don't see any reason for this not to be the right answer. Thanks for the help everyone :)