Question

integral (u^2)/[sqrt -4(u^2)+4]^3 du
can I factor out a (-1/4) outside the integral?

Answer 1

\[\int\limits_{}^{}\frac{u^2}{\sqrt{(-4u^2+4)^3}} du\]
Is this correct?

Answer 2

the sqrt is all raised to the third powr

Answer 3

Yeah that is 3 only and not 2..

Answer 4

\[\int\limits_{}^{}\frac{u^2}{(\sqrt{-4u^2+4})^3} du\]

Answer 5

This looks like you would need a trig sub either way.

Answer 6

yea, I'm prepping for a trig sub, but I want my form u^2 +4

Answer 7

but since the (-4) lies within the sqrt, I'm not remembering my algebra if I can bring it out or not

Answer 8

\[(\sqrt{-4u^2+4})^3=(\sqrt{4(-u^2+1)})^3=(\sqrt{4} \sqrt{-u^2+1})^3=(\sqrt{4})^3(\sqrt{-u^2+1})^3\]
So this part right here is the part that you need help on?

Answer 9

yes

Answer 10

looks like rt 4^3 is a coefficient I can factor when you write it like that

Answer 11

Yes. I just used some law of exponents stuff.
Like I used
\[((ab)^\frac{1}{n})^b=(a^\frac{1}{n}b^\frac{1}{n})^b=a^\frac{b}{n} b^\frac{b}{n}\]

Answer 12

some tutors at school were just popping out the -4, but it didn't seem I could do that since it was within the root.

Answer 13

You mean bring it outside the integral?

Answer 14

yes

Answer 15

Yes you can bring constant multiplies outside the integral

Answer 16

\[\frac{1}{(\sqrt{4})^3} \int\limits_{}^{}\frac{u^2}{(\sqrt{-u^2+1})^3} du\]

Answer 17

I mean that it's incorrect to go from the first equation we started with to just bringing (-1/4) outside the integral. Since (-4) is inside the sqrt, it cannot be brought outside teh integral as (-1/4).

Answer 18

When you write it as 1/(sqrt 4)^3 it makes more sense.

Answer 19

And of course that can be simplified