Indefinite integrals? Not sure how different they are from regular integrals. I'm really bad at anti derivatives so if they come up can someone explain how you found the antiderivative as well?

integral of 6x^2(2x^3 - 5)^5 dx

Question

Indefinite integrals? Not sure how different they are from regular integrals. I'm really bad at anti derivatives so if they come up can someone explain how you found the antiderivative as well?

integral of 6x^2(2x^3 - 5)^5 dx

anonymous · Answer

substitute 2x^3-5 = t

anonymous · Answer

6x^2 dx = dt

anonymous · Answer

$$\int_{10}^{\infty}\frac{6x^2}{(2x^3-5)^5}dx$$??
you said it was "improper" but did not specify the limits of integration

anonymous · Answer

I don't know what you mean satellite73

anonymous · Answer

oh nvm you said "indefinite" ignore me i don't know how to read

anonymous · Answer

what @divu.mkr said, put $u=3x^2-5, du =6x^2dx$ and you are done in one step

anonymous · Answer

Since it is indefinite and there are no limits to fix, that's all I need to finish the problem?

freckles · Answer

and he meant u=2x^3-5 :p

anonymous · Answer

write all in terms of $u$

anonymous · Answer

yeah that one $u=2x^3-5$ sorry

anonymous · Answer

Ah okay. Thank you very much! Teacher gave me a packet on Friday and I've been slowly working through them for a couple days.

freckles · Answer

you can do the integration part right?
not the sub part I guess?
I'm just asking so I know what part is the part you find difficult

anonymous · Answer

I think so. I'm working through it right now to see what I get

freckles · Answer

post here if you want it checked

anonymous · Answer

Okay I got (2x^3 - 5)^6 / 6

freckles · Answer

+C 
and yah

anonymous · Answer

uggggghhh I always forget +C
Alright thanks to those who replied!