Question

Find the integral of t^3(t-4)^(1/2)dt

Answer 1

u=t-4

Answer 2

So du=dt

Answer 3

yes

Answer 4

Ok then after I substituted some of the things, I got (u+4)^3u^(1/2)du but then I got stuck there...

Answer 5

expand\((u+4)^3\)

Answer 6

\[\large \int\limits (u+4)^3\sqrt u \; du\]Just so it's easier to read D:

Answer 7

Ok so did you get u^3+12u^2+48u +64?

Answer 8

that is correct...now distribute the \(\sqrt{x}\)

Answer 9

\[\sqrt{u}\]

Answer 10

then integrate term by term

Answer 11

I got 2u^(9/2)/9+4u^(7/2)/7+96u^(5/2)/5+64u(3/2)/3+c, and all I need to do is change the u's to (t-4)?

Answer 12

Wait the second term should be 24u^(7/2)/7 and the last should be 128 u^(3/2)/3

Answer 13

correct

Answer 14

Thank you :)