Integrate (x^3)/sqrt(4x^(2)+9)^(3): Steps & Tips

[SJB3415]

How do you integrate: (x^3)/sqrt(4x^(2)+9)^(3) dx

I have done integration with sqrts before but not when it is cubed up the square root...Any suggestions?

 

[planck42]

How can you remove the $$x^3$$ term?

 

[FanofAFan]

You could try pluging in the problem in Wolfram|alpha...

I actually did for you... follow the link, they work it out in complete steps
http://www.wolframalpha.com/input/?i=(x^3)/sqrt(4x^(2)+9)^(3)

 

[n!kofeyn]

just to make sure, is this your problem?
$$\int \frac{x^3}{(4x^2+9)^{3/2}} \,dx$$

i see two possible approaches to this problem. although this doesn't look like a typical u-substitution problem, try making one. hint: you usually let the u be the thing inside a composition of functions, especially if it's underneath a square root. so, try letting u=4x²+9. then you have the problem of accounting for the x³. break x³ up into x²x. then you'll need to find x² in terms of u. after all this, it should be a simple integral to finish.

the other method (that i haven't worked out) would be to try a trig-substition. in this case it would be $x=\frac{3}{2}\tan\theta$. like i said, i didn't take the time to work this out, but this should work if you can finish the trig integral you'll be left with after making the substitution.

You could try pluging in the problem in Wolfram|alpha...

I actually did for you... follow the link, they work it out in complete steps
http://www.wolframalpha.com/input/?i=(x^3)/sqrt(4x^(2)+9)^(3)

no where in that link did they actually work this problem out in complete steps. wolframalpha is only helpful for checking your answers.

 

[FanofAFan]

where it says on the right side in orange letter, "Show sets" under integration

 

[n!kofeyn]

where it says on the right side in orange letter, "Show sets" under integration

my apologies, and i stand corrected. although the steps could be seen as confusing. they actually make the substitution that i mentioned above, but they do so using two separate substitutions.