I can't figure out how to do this but Ive been assured its easily doable.

Calculate: (int) 2x(4x+7)^{8 }dx

So I'm assuming we try to get 4x on its own and find 2x in terms of u?

A solution to this would help me heaps here I just cant see it!

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- Jan 5th 2013, 02:39 PMcamjerlamsHelp with a U substitution problem.
I can't figure out how to do this but Ive been assured its easily doable.

Calculate: (int) 2x(4x+7)^{8 }dx

So I'm assuming we try to get 4x on its own and find 2x in terms of u?

A solution to this would help me heaps here I just cant see it! - Jan 5th 2013, 02:48 PMskeeterRe: Help with a U substitution problem.
$\displaystyle \int 2x(4x+7)^8 \, dx$

$\displaystyle u = 4x+7$

$\displaystyle x = \frac{u-7}{4}$

$\displaystyle dx = \frac{du}{4}$

$\displaystyle \int 2x(4x+7)^8 \, dx$

$\displaystyle \int \frac{u-7}{8} \cdot u^8 \, du$

$\displaystyle \frac{1}{8} \int u^9 -7u^8 \, du$

can you finish? - Jan 5th 2013, 03:02 PMcamjerlamsHelp with a U substitution problem.
Thanks! Im not sure if I can Im really having trouble with integration. Would appreciate the rest.

- Jan 5th 2013, 03:44 PMskeeterRe: Help with a U substitution problem.
If you're at the level of doing integrals by substitution, then this should be basic stuff.

what is the antiderivative of $\displaystyle u^9$ ?

what is the antiderivative of $\displaystyle u^8$ ? - Jan 5th 2013, 03:53 PMcamjerlamsRe: Help with a U substitution problem.
I havent studied math for about 10 years so i struggle with the basic stuff.

so antiderivative of u^9=1/10u^10

and for u^8=1/9u^9

so 1/8 (1/10 u^10 - 7/9 u^9) ?? - Jan 5th 2013, 03:56 PMskeeterRe: Help with a U substitution problem.
- Jan 5th 2013, 03:59 PMcamjerlamsRe: Help with a U substitution problem.
Thanks skeeter, big help, I know I need practice

- Jan 5th 2013, 04:47 PMDevenoRe: Help with a U substitution problem.
note you don't have to resort to u-substitution, you COULD just multiply out (4x+7)

^{8}, and then multiply every term by 2x, and then integrate term-by term. this is kind of "the hard way", u-substitutions are meant to make your life EASIER (short-cuts are good, right?).

linear substitutions u = ax+b are always "do-able", we just have a "fudge factor" of a constant to worry about:

du = a dx, so

dx = (1/a) du <--take the 1/a outside the integral sign. - Jan 6th 2013, 12:53 AMcamjerlamsRe: Help with a U substitution problem.
so, 1/8((1/10u^10) - (7/9u^9))

=1/8((4x+7)^10/10 - 7/9(4x+7)^9) - Jan 6th 2013, 03:15 AMDevenoRe: Help with a U substitution problem.
don't forget to add the constant of integration....