Question

U Substitution Question

Answer 1

\[\int\limits_{}^{}\frac{ x }{( 8x+5)^2 } where u=8x+5\]

Answer 2

u=8x+5
I have it calculated out like so, so far.

Answer 3

\[\int\limits_{}^{} \frac{ x }{ u^3 } dx\]

Answer 4

u = 8x + 5

du/dx = 8

du = 8*dx

du/8 = dx

dx = du/8

Answer 5

u = 8x + 5

u - 5 = 8x

8x = u - 5

x = (u - 5)/8

Answer 6

right i have that part figured out. but cant seem to put things together.

Answer 7

I know im sopost to sub x in right? but then do I sub U back into the old x placeholder again?

Answer 8

so what this means is that

\[\large \int\frac{ x }{( 8x+5)^2 } * dx \]

turns into

\[\large \int\frac{ \frac{u-5}{8} }{u^2 } * \frac{du}{8} \]

and that simplifies to

\[\large \int\frac{ u-5 }{64u^2 } du \]

Answer 9

so what do i do with the u at that point? Just sub back in 8x+5?

Answer 10

or do I take the anti-derrivative from there first then sub u back in

Answer 11

you need to integrate first

Answer 12

in terms of u

Answer 13

after you've done that, you would then sub 8x+5 back in for u

Answer 14

k think ive got it as I keep forgetting to take the anti-deriv

Answer 15

\[\large \int\frac{ u-5 }{64u^2 } du \]

\[\large \frac{1}{64}\int\frac{ u-5 }{u^2 } du \]

\[\large \frac{1}{64}\int(\frac{ u}{u^2}-\frac{5}{u^2 }) du \]

\[\large \frac{1}{64}\int(\frac{1}{u}-\frac{5}{u^2 }) du \]

\[\large \frac{1}{64}\left[\int\frac{1}{u}du-\int\frac{5}{u^2 } du\right] \]

\[\large \frac{1}{64}\left[\int\frac{1}{u}du-\int5u^{-2}du\right] \]

I'm sure you see what to do from here

Answer 16

yeah i do, and thanks

Answer 17

yw