Question

pleasee help!! stuck on this question!

Decide on what substitution to use, and then evaluate the given integral using a substitution.

x(x-6)^1/4 dx

Answer 1

So far I did

U=x-6

du= 1 dx

so dx= 1du ?

Answer 2

Yup

Answer 3

your x wont cancel like you are probably expecting, so you'll need to solve for it in terms of u.

Answer 4

so I got then 1 (integral symbol) u^1/4 +1 du

=1 u^5/4/(5/4) +c
=1 (x-6)^5/4/(5/4) + C

Answer 5

where did I go wrong :/?

Answer 6

\[\int\limits_{}^{}(u-6)(\sqrt{u})du\]

Answer 7

where did the square root come from?

Answer 8

im sorry I read the 1/4 as 1/2

Answer 9

oh okay no worries. so its ∫(u−6)(u)du

Answer 10

an I plug in the "u" ?

Answer 11

yes with u^(1/4)

Answer 12

so would it be (x-6)(x-6)^1/4 du ?

Answer 13

then multiply the u's --> u^(5/4)-6u^(1/4)

Answer 14

Then split the integral

Answer 15

so -6u^3/2