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## Homework Statement

The definite integral of from 0 to 1 of ∫ (r

^{3})dr/sqrt(4+r

^{2})

## Homework Equations

∫udv = uv - ∫vdu

∫du/sqrt(a

^{2}- u

^{2}) = arcsin(u/a) + C

∫du/(asqrt(a

^{2}- u

^{2})) = (1/a)arcsec(u/a) + C

## The Attempt at a Solution

I made u = (4+r

^{2})

^{-1/2}

because I thought it easier to get it's derivative, rather than integral by making it dv.

du = -r(4+r

^{2})

^{-3/2}dr

dv = r

^{3}dr

v = (1/4)r

^{4}

I've set up the equation as

= uv -∫vdu

= r

^{4}/(4(4+r

^{2})

^{1/2}) + ∫r

^{3}dr/(4+r

^{2})

^{3/2}

This seems somewhat similar to the the second equation I put in "Relevant Equations" (where I could put a two in the numerator and multiply the integral by (1/2), but both top and bottom are to the power of 3 (which I don't know how to get rid of. AND the bottom is not a

^{2}- u

^{2}but a

^{2}+ u

^{2}.

If somebody could please help me understand this I would be greatly appreciative. I'd like to be able to master this type of integration, but I seem to have hit a wall.