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\[\int\limits_{?}^{?} x^3(x^4-10)^{45}dx \]
by making the substitution that u=x^4-10

ok u substituted u
now find what is dx in terms of du

\(u=x^4-10,\quad du=4x^3dx\)
\(\int\limits x^3(x^4-10)^{45}dx=\frac14\int u^{45}du\)

Don't forget to substitute (x^4 - 10) back in for you!

lol u*

-10, sorry bout that. Shouldn't make much difference.

|dw:1358443764659:dw|\[\huge \rightarrow \int\limits (u)^{45}\left(\frac{1}{4}du\right)\]

now do i substitute u and du in for the integral? and I need it to be in respect to x not u

can i always divide how you did to get the 1/4?

So I had to get the 4 OFF of it.

1/4x^3 dx?

or is it replaced as a C?

So we need to balance it..?

Okay that makes sense.

what comes after balancing it though..?

1/4(x^4-10)^(45)*x^3
?

oh okay I see how you substituted it.

now i need to antidifferentiate?

Yes good good c:

umm..what part of that though? (x^4-10)^{45} ? and x^3dx

Did you make the substitution?
We don't integrate until we've changed it over to U.

don't worry about all that X garbage. We got rid of that with our substitution.

yeah i got that part.

yeah im having trouble integrating with respect to u... can you explain that more clearly

yeah that makes sense :) i forgot that formula

got it :)

yay team c:

wait but where does the x^3dx go then?...

You don't have to back substitute the DU.
The DU disappears as a part of the integration process.

and thats why it becomes +C?