The integral of 1/(x^2+9)^(3/2)dx

Help!

Question

The integral of 1/(x^2+9)^(3/2)dx

Help!

anonymous · Answer

Yes, but this problem deals with the the In of a function...

anonymous · Answer

the last one wasn't correct, not trolling. Just trying to help. i reposted on your old question

anonymous · Answer

What do you mean it wasnt correct? I said tan^5x/5 +c

anonymous · Answer

your post said tan^2x/2+c, looks like just a typo

anonymous · Answer

Lol yeah it must have been a typo :P Sorry and thank you :)

anonymous · Answer

So do you have any advice for this problem?

anonymous · Answer

sorry i've got no luck with this one yet...>.<

anonymous · Answer

Okay...Have you tried Anti-log?

anonymous · Answer

you mean like 1/x =ln(x)*x'?

anonymous · Answer

Yeah....Its the first one. :)

anonymous · Answer

wolfram got the answer of $$\frac{ x }{ 9\sqrt{x^2+9}}$$...i havent been able to get that

anonymous · Answer

Yeah I saw that on my calculator  but I havent been able to get to that answer either :/

anonymous · Answer

Okay nvm dude...I solved it :)

anonymous · Answer

nice, what'd you use?

anonymous · Answer

I set x=Squ(9)sectheta and dx=Squ(9)(sectheta)(tantheta)(dtheta)

anonymous · Answer

I used Squ(x^2-a^2) which means x=(a)sectheta

anonymous · Answer

And yeah :P I did a lot of work after that. But thats how I started and I finished :)

anonymous · Answer

ahhh nice

anonymous · Answer

Yep :) Thanks for the help on the last problem :P