Question

Integration by Parts

Answer 1

integrate \[\frac{ x^2 }{ \sqrt{x^2+14} }\]

Supposed to use integration by parts method, but stuck. Any help appreciated!

Answer 2

any idea how to start

Answer 3

I thought of using x^2 as u, and 1/sqrt(x^2+14) as dv.

dv then kinda looks like an inverse tangent but not sure.

Answer 4

It does on second thought look like a inverse sinh, but I've not learned about them before so not sure.

Answer 5

Well you can pretty much always use the regular trig functions you're used to instead of the hyperbolic trig functions, it just ends up being a little more difficult sometimes. For this particular case it looks like this might be related to inverse tangent, I'm specifically thinking of this identity: \[\Large \tan^2 \theta + 1 = \sec^2 \theta\]

Answer 6

Hmmm...how do I relate that to \[ \frac{ 1 }{ \sqrt{x^2+14}}\] ?

Answer 7

Do you mean to say I can directly substitute it as u, where u = that identity?

Answer 8

Well it says you're supposed to use integration by parts, so I'm not entirely sure where that will come in. However I can safely say that if you were to plug anything in right now you could try \[\Large x= \sqrt{14} \tan \theta\] Really what will make it more clear is if you look at the bottom part, factor out the 14 from the whole thing: \[\Large \frac{1}{\sqrt{x^2+14}} = \frac{1}{\sqrt{14}} \frac{1}{\sqrt{\frac{x^2}{14}+1}}\]

Answer 9

Ah I see, this is some high level stuff here lol, will try and do it slowly now. Will update later, thanks!

Answer 10

Yeah definitely! You get used to it though, it's really not so bad as it seems haha.

Answer 11

another way to compute that integral is given bythis substitution:
\[\sqrt {{x^2} + 14} = x + t\]
where t is the new variable of integration.
Squaring both sides, you shoul get this:
\[x = \frac{{14 - {t^2}}}{{2t}}\]
now, we can start from there

Answer 12

oops..should*