Question

∫x^3dx/^5rad(x^4+6)

Tried to write it best way I could. Having trouble choosing which to use as my u substitution, would x^3 be good to use?

Answer 1

When I do x^3 as my u sub I get stuck with the ugly number on the bottom of the fraction and I don't know what to do with it

Answer 2

\[\int\limits_{}^{}\frac{ x^{3} }{ \sqrt[5]{x^{4} + 6} }dx\]??

Answer 3

yes!

Answer 4

sorry I didn't know how to write it like that on here

Answer 5

A lot of the time your natural choice for a u-sub is going to be something inside of a radical or in the denominator of a fraction. Of course it's not always like that, but it's common. So in this case, your u-sub should be \(u =x^{4} + 6\)

Answer 6

so when I try that one, my u is x^4+6 and my du is equal to 4du = x^3dx?

Answer 7

\(du = 4x^{3}dx\)

Answer 8

oh I don't have to divide both sides by 4 to get rid of it?

Answer 9

Well, you had the 4 in the correct spot anyway if you were to divide both sides by 4. But I see what you tried to do. But yes, your idea is perfectly fine \(\frac{du}{4} = x^{3}dx\)

Answer 10

*incorrect
wow, fail me.

Answer 11

thanks! and I am not supposed to find the derivative of x^3 and make it 3x^2 correct?

Answer 12

Nah, there's no reason to. If you can get a u and a du that lets you turn your integral into only u's, du, and constants, then you don't need to do anything else fancy. Given \(\frac{du}{4} = x^{3}dx\) you have
\[\int\limits_{}^{}\frac{ du }{ 4\sqrt[5]{u} }\] which is perfectly fine to integrate now.

Answer 13

Great! thanks! it it correct if i turn the bottom part of the fraction to 4 * 1/5x^(4/5)

Answer 14

if you don't like radicals, you could also try :
\[\rm u^5 = x^4+6\]

Answer 15

ok, thank you! I will try it