Question

how the heck is this integrated??

this is my working.. did something go wrong?

http://flipapicture.com/uploaded_images/115651_phto.jpg

Answer 1

split the fraction, then its just 2 power rules that i see

Answer 2

you seem to have worked it down to the power split :)

Answer 3

do a substitution: u = 4x+1, what is dx in terms of du?

Answer 4

uhmmmm i don't know? sorry haha i mean, I'm not familiar with the verbal terms they're called, I learn by examples ><

Answer 5

well, in order to find du or dx, we just take the derivative of u-f(x)

Answer 6

isn't dx in terms of du just dx/du?

Answer 7

u = f(x)

Answer 8

u = 4x + 1

du = 4 dx

Answer 9

bu but what about the -2 power?

Answer 10

u^-2 .. what about it?

Answer 11

\[\int au^2+bu^{-2}~du\]

Answer 12

i don't know how to integrate a negative power

Answer 13

that's like, outside a thing

Answer 14

have you covered a power rule?

Answer 15

i mean, x^-2 i can do, but if its like (___)^-2 i think i can't integrate it the normal way..

Answer 16

x^-2 = (x)^-2

Answer 17

if i can't integrate (x-2)^2 without expanding, how can I integrate (x-2)^-2?

Answer 18

thats why we are changeing the variable .... we are essentialy moving the integral into a more workable space.

Answer 19

if we let u = 4x+1 then we convert the problem from the xy plane into the uv plane

Answer 20

well, the uy plane

Answer 21

ohh... so once the bracket is subbed into an unknown, and i integrate, then how do i re-equate the bracketed back into the equation?

Answer 22

u still equals 4x+1

Answer 23

but the inside of u isn't integrated

Answer 24

lets work the substitution

\[\int a(4x+1)^2+b(4x+1)^{-2}~dx\]
replace 4x+1 with u, since they are equal
\[\int a(u)^2+b(u)^{-2}~dx\]
replace dx with its du equivalent

4dx = du when we take the derivatives: dx = du/4
\[\frac14\int a(u)^2+b(u)^{-2}~du\]

Answer 25

okay

Answer 26

what if i don't take out the 4?

Answer 27

what do you mean?

Answer 28

why did you put a 1/4 outside if you already subbed out the entire 4x+1?

Answer 29

du/4 = 1/4 du

1/4 is just a constant that can be pulled outside the integration

Answer 30

\[\int [a(u)^2+b(u)^{-2}]~\frac{du}{4}\]
\[\int \frac14[a(u)^2+b(u)^{-2}]~du\]
\[ \frac14\int [a(u)^2+b(u)^{-2}]~du\]

Answer 31

you can pull out a 1/2 as well to reduce a and b :) but thats immaterial

Answer 32

using power rule
\[\frac14\int [a(u)^2+b(u)^{-2}]~du=\frac14[\frac13u^3- bu^{-1}]\]

Answer 33

and we know what u is equal to if we want to sub it back in ... otherwise we can just adjust the interval of integration from x space to u space

u = 4x + 1, when x=0, u=1
when x=1, u=5

Answer 34

okay..

Answer 35

\[\int\limits_{x=0}^{x=1} a(4x+1)^2+b(4x+1)^{-2}~dx\]
since u = 4x+1, u substitution converts this to x = (u-1)/4
\[\large \frac{1}{4}~~\int\limits_{(u-1)/4=0}^{(u-1)/4=1} a(u)^2+b(u)^{-2}~du\]
\[\large \frac{1}{4}~~\int\limits_{u=1}^{u=5} a(u)^2+b(u)^{-2}~du\]

Answer 36

ah okay thank you ^^

Answer 37

i was wondering, would you know how to answer this?

http://openstudy.com/study#/updates/54394df9e4b01d33976580f1