Question

indefinite integral of t^2 (t+2)^(1/2)

Answer 1

you can do integration by parts, or add a magic zero a munch of times.

Answer 2

By magic zero I mean,
\[\int\limits_{ }^{ }t^2\sqrt{t+2}+(4t+4)\sqrt{t+2}-(4t+4)\sqrt{t+2}dt\]\[\int\limits_{ }^{ }(t^2+4t+4)\sqrt{t+2}-(4t+4)\sqrt{t+2}dt\]\[\int\limits_{ }^{ }(t+2)^2\sqrt{t+2}-(4t+4)\sqrt{t+2}dt\]\[\int\limits_{ }^{ }(t+2)^{5/2}-(4t+4)\sqrt{t+2}dt\]

Answer 3

Doing the magic zero again.
\[\int\limits_{ }^{ }(t+2)^{5/2}-(4t+4)\sqrt{t+2}-2\sqrt{t+2}+2\sqrt{t+2}~dt\]\[\int\limits_{ }^{ }(t+2)^{5/2}-(4t+4+2)\sqrt{t+2}+2\sqrt{t+2}~dt\]\[\int\limits_{ }^{ }(t+2)^{5/2}-(4t+8)\sqrt{t+2}+2\sqrt{t+2}~dt\]\[\int\limits_{ }^{ }(t+2)^{5/2}-4(t+2)\sqrt{t+2}+2\sqrt{t+2}~dt\]

Answer 4

\[\int\limits_{ }^{ }(t+2)^{5/2}-4(t+2)^{3/2}+2\sqrt{t+2}~dt\]

Answer 5

interesting trick tho

Answer 6

So if you haven't learned integration by parts, this is pretty much the same thing, but you would be doing Algebra instead of integration. you are still increasing the exponents of (t+2) as you would, when doing integration by parts.

Answer 7

@gbblue you should be able to take it from here.

Answer 8

lol

Answer 9

it is just like integrating csc(x), but substituting 2u instead of x, and then rewriting the bottom with sin(a+a) rule, and on....

Answer 10

@gbblue do you feel comfortable with any of the posted approaches?
Or do you want to do integration by parts instead.
or do you just want to get a bad grade?

Answer 11

Lol.@idku for you last comment.:P

Answer 12

I wanted to leave the last comment, or second to last to gbblue , but he/she doesn't seem to want to do the problem.

Answer 13

I am not coming back here again....

Answer 14

sorry I was getting coffee, but yes this was very helpful and I can take it from here

Answer 15

@gbblue check my method too

Answer 16

Finally.... it's alright. I hope you have a good coffee for the least oart.

Answer 17

@Princer_Jones I do not quite understand your method

Answer 18

it is integration by trigonometric substitution