# Help with a U substitution problem.

• Jan 5th 2013, 02:39 PM
camjerlams
Help 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 PM
skeeter
Re: Help with a U substitution problem.
$\int 2x(4x+7)^8 \, dx$

$u = 4x+7$

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

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

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

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

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

can you finish?
• Jan 5th 2013, 03:02 PM
camjerlams
Help 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 PM
skeeter
Re: 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 $u^9$ ?

what is the antiderivative of $u^8$ ?
• Jan 5th 2013, 03:53 PM
camjerlams
Re: 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 PM
skeeter
Re: Help with a U substitution problem.
Quote:

Originally Posted by camjerlams
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) + C

was that so difficult?

now finish it by back-substituting (4x+7) for u ...
• Jan 5th 2013, 03:59 PM
camjerlams
Re: Help with a U substitution problem.
Thanks skeeter, big help, I know I need practice
• Jan 5th 2013, 04:47 PM
Deveno
Re: 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 AM
camjerlams
Re: 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 AM
Deveno
Re: Help with a U substitution problem.
don't forget to add the constant of integration....