Question

integrate (sqroot x)/x-1)

Answer 1

Let me check my answer then I'll post it.

Answer 2

Okay, you have:
\[\int\limits \frac{\sqrt x}{x-1}dx\]
Make the substitution:
\[u=\sqrt{x} \implies u^2=x \implies 2u du=dx\]
Making that change of variables we get:
\[\int\limits \frac{\sqrt{x}}{x-1}dx=\int\limits \frac{u}{u^2-1}(2u du)\]
From here we can simplify this:
\[2 \int\limits \frac{u^2}{u^2-1}du=2 \int\limits du+2 \int\limits \frac{du}{u^2-1}\]
Doing long division, then factoring the right integral we have:
\[2u+2 \int\limits \frac{du}{(u+1)(u-1)}=2u+2(\frac{1}{2})\int\limits \frac{du}{u+1}-2(\frac{1}{2})\int\limits \frac{du}{u-1}\]
Making the substitution p=u+1 s=u-1 dp=ds=du
We get:
\[2u+\int\limits \frac{dp}{p}-\int\limits \frac{ds}{s}=2u+\ln|p|-\ln|s|=2u+\ln|u+1|-\ln|u-1|+C\]
Making appropriate substitutions and then we arrive at:
\[2\sqrt{x}+\ln|\sqrt{x}+1|-\ln|\sqrt{x}-1|+C\]

Answer 3

How did you simplify the part after you wrote "
From here we can simplify this:"?

Answer 4

I just did long division. When you divide u^2 by u^2-1 you get 1+1/(u^2-1), I just dropped the "implied" ones.

Answer 5

alright, so then how were you able to pull a (1/2) out in the next step?

Answer 6

Partial fractions :P I split up the (blah)/((u+1)(u-1)) into something in the form:
\[\frac{A}{u+1}+\frac{B}{u-1}\]
Got a common denominator and then solved for A and B. THEN integrated (when you solve the constants I get the stipulation that A=-B and 1=A+B, solving these we get A=1/2 and b=-1/2; thats how I pulled them out, I just didn't wanna drop the two so you could see where that cancellation occured.)

Answer 7

YAY!!! I finally got it! Thanks so much! You have been so VERY HELPFUL! It takes me so long to get calculus. I wish I could give you more medals...Thank you again ;D

Answer 8

No problem :D Good luck xDD Need anymore help just post them in the feed

Answer 9

;)