how do you integrate this?

Question

find the integral of 2x
-------------- dx
(2x-1)squared

using the substitution u = 2x-1

any help is greatly appreciated.

Answer 1

i don't understand the way you
have written down the function

say that you want to integrate
2x/(2x-1)
put u=2x-1,>>>x=(u+1)/2,
dx=du/2,
int{2x/(2x-1)}dx
=int{(2*(u+1)/2)/u}*du/2
=1/2*int{(u+1)/u}du
=1/2*int{1+1/u}du
=1/2[u+lnu]+C
but u= 2x-1
=1/2[2x+ln(2x-1)-1]+C
where C is a constant

i hope that this helps

Answer 2

So you want to solve

Integral ( (2x-1)^2 ) dx

Let u = 2x - 1
Then du = 2 dx
and du/2 = dx

Therefore, all you have to do is substitute (2x-1) with u and dx with what we just solved for.

Integral (u^2) du/2

We should pull out ALL constants, to get

1/2 * Integral (u^2) du

This is a simple integration using the reverse power rule.

1/2 (u^3/3) + C

Substitute u back in, to get

1/2 [(2x-1)^3]/3 + C

And pulling out all constants:

1/6 * (2x-1)^3 + C

Answer 3

u = 2x-1, du/dx = 2 => du = 2 dx => dx = (1/2) du

Integral of 2x/(2x-1) dx
= Integral of u+1/u dx
= Integral of u+1/u (1/2) du
= Integral of u+1/(2u) du
= Integral of (1/2) + (1/2)u^-1 du
= (1/2)u + (1/2)ln u + C
= (1/2)(2x-1) + (1/2)ln(2x-1) + C

Answer 4

add and subtract 1 to the numerator
integral of 2x-1+1/(2x-1)^2
int(dx/(2x-1)+dx/(2x-1)^2
1/2log(2x-1)-1/2(2x-1)+C

Answer 5

By going down the SU bar, getting drunk and copying someone elses hard work.

Make mine a snakebite!

Answer 6

I dont becasue I have an axe