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C4 maths help please!!

I have some questions to do on integration by parts and I have NO idea what I'm doing. One of the questions is "find the integral of 3x(root(x-1))dx"

Somebody help please!
(edited 7 years ago)
if u = 2x-1
then the top of the fraction 2x would be u+1

& du/dx = 2

giving you

1/2 * integral of [ (u+1) / (u^2) ]

which is 1/2 * the integral of [ u + u^-2 ]

....
(edited 7 years ago)
Original post by kaagyu
I have some questions to do on integration by parts and I have NO idea what I'm doing. The first question is "By using the substitution u=2x-1, or otherwise, find the integral of 2x/(2x-1)^2"

Somebody help please! I've substituted in u and that's as far as I can get.


OK but you also need to change your integral to be du - you can't integrate a function of u wrt x.

u = 2x - 1

du/dx = 2 so du = 2 dx now substitute for dx after rearranging
Original post by kaagyu
I have some questions to do on integration by parts and I have NO idea what I'm doing. The first question is "By using the substitution u=2x-1, or otherwise, find the integral of 2x/(2x-1)^2"

Somebody help please! I've substituted in u and that's as far as I can get.


okay
so u=2x-1
therefore 2x=u+1
now since the original is dx we have to convert to du

so differentiate u=2x-1 in respect to x
du/dx = 2
therefore du=2dx
so dx=du/2

thus the integral of 2x/(2x-1)^2 dx
becomes

(u+1)/(2(u^2)) du

which simplifies to
= (u/(2u^2)) +(1/(2u^2)) du

= 0.5u^-1 + 0.5u^-2 du
now integrate that bit

=0.5ln(u) -0.5u^-1 + c
now change all u to 2x-1
=0.5ln(2x-1) - 0.5(2x-1)^-1 + c
Original post by suhaylpatel786
okay
so u=2x-1
therefore 2x=u+1
now since the original is dx we have to convert to du

so differentiate u=2x-1 in respect to x
du/dx = 2
therefore du=2dx
so dx=du/2

thus the integral of 2x/(2x-1)^2 dx
becomes

...


Please read the guidelines in posting .... we don't post full solutions in Maths - just hints :smile:
Original post by kaagyu
I have some questions to do on integration by parts and I have NO idea what I'm doing. One of the questions is "find the integral of 3x(root(x-1))dx"

Somebody help please!

First take the 3 out
giving

3 * the integral of [ x * root(x-1) ] dx

then let u = x-1
du/dx = 1
giving
3 * the integral of [ (u+1) * root(u) ] du

expand the integral
giving

3 * the integral [ u^3/2 + root(u) ] du
(edited 7 years ago)

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