Question

need help with integral problems

Answer 1

\[\int\limits_{0}^{1}\sqrt{x}*(1-x)\]

Answer 2

and this one too \[\int\limits_{?}^{?} \frac{ x+4 }{ (x^2+8x- 7)^2}\]

Answer 3

so just multiply the x^1/2 for the first one and what about the second one?

Answer 4

hello?

Answer 5

Like i said, for the first one, all you have to do is distribute x^(1/2) into he parenthesis. For the second problem, do u-sub for the second one.
Let u=x^2+8x-7
du=2x+8 dx
also, can be written as du=2(x+4)dx

Answer 6

\[\int\limits_{}^{}\frac{x+4}{(x^2+8x-7)^2}dx=\frac{1}{2}\int\limits_{}^{}\frac{2(x+4)}{(x^2+8x-7)^2}\]
\(\large u=x^2+8x-7\) and \(du=(2x+8)dx\)

Answer 7

@El_Arrow clear?

Answer 8

why do you have the 1/2 out front?

Answer 9

All I'm doing is wishing 2 to be in the integral to make it look like the du= 2(x+4)dx,
but i'd have to factor out 1/2 . Anything you want to multiply in the integral must have another number than can make it go back to normal

\(\LARGE \frac{1}{2}\int\limits_{}^{}\frac{2(x+4)}{(x^2+8x-7)^2}\)=\(\LARGE \int\limits_{}^{}\frac{1}{2}\frac{2(x+4)}{(x^2+8x-7)^2}\)=\(\LARGE \int\limits_{}^{}\frac{(x+4)}{(x^2+8x-7)^2}\)

When half gets multiplied back into the integral, it gets cancelled by the two. ALL of these three integrals are the same except a couple of algebra was done to make it look easier to proceed with the U-Sub.

Answer 10

I'm forgetting the dx.

Answer 11

My bad.

Answer 12

okay so your left with

Answer 13

what do you do from there?

Answer 14

Nopes. What happened to the 1/2 and 2(x+4) ? YOU need 2(x+4) because that's what you want to replace with the du.