Question

Integrate:

Answer 1

\[\LARGE \int x^2~\sqrt{2+x}~dx\]

Answer 2

2+x=u

it's kinda tricky.

Answer 3

Substitute \(u=x+2\), then \(du=dx\) and \(x^2=(u-2)^2\).

Answer 4

I wonder if you could solve this by making the substitution \[x=2\tan^2 \theta\] lol

Answer 5

This is the type of u-sub that I never understood.

So we end up with
\[\LARGE \int (u-2)^2 ~\sqrt{u}~du\]

Answer 6

Yes, does it make sense how we get there though?

\[\int\limits x^2 \sqrt{2+x} \ dx\]
\[u=2+x\]\[u-2=x\]\[\frac{du}{dx}=1\]\[du=dx\]
\[\int\limits(u-2)^2\sqrt{u} \ du\]\[\int\limits (u^2-2u+4)u^{1/2}\ du\]\[\int\limits u^{5/2}-2u^{3/2}+4u^{1/2}\ du\]

Answer 7

I totally squared that wrong, should be (u^2-4u+u) lol.

Answer 8

Ohh, haha, I see what you were trying to do tho, thank you for the explanation, really helped! :)

Answer 9

Glad I could help!

Answer 10

Still going... WOOHOO