# Integration by substitution

• Feb 5th 2008, 07:25 PM
simsima_1
Integration by substitution
$
\int \frac{2x-1}{(4x^2-4x)^{2}} dx\

$

In this case what would u be? that the integral would be

$

\frac{1}{16} \, \int u^{-2} du

$

At first glance i thought it would just be
$
u=4x^2-4x
$

but obviously thats not it.
• Feb 5th 2008, 07:39 PM
ThePerfectHacker
Quote:

Originally Posted by simsima_1
$
\int \frac{2x-1}{(4x^2-4x)^{2}} dx\

$

In this case what would u be? that the integral would be
.

$(4x^2-4x)^2 = 16(x^2 - x)$.
Let $u=x^2-x$.
• Feb 5th 2008, 07:56 PM
simsima_1
Quote:

Originally Posted by ThePerfectHacker
$(4x^2-4x)^2 = 16(x^2 - x)$.
Let $u=x^2-x$.

....dumb question ... but how did u get that? the 16(x^2-x)??
• Feb 5th 2008, 08:00 PM
ThePerfectHacker
Quote:

Originally Posted by simsima_1
....dumb question ... but how did u get that? the 16(x^2-x)??

$(4x^2-4x)^2 = (4(x^2-x))^2 = 4^2(x^2-x)^2 = 16(x^2-x)^2$

EDIT: Mistake fixed.
• Feb 5th 2008, 11:47 PM
Lopared
$(4x^2-4x)^2=(4(x^2-x))^2$
if $x^2-x = u$ then $(2x-1)dx=du$ so you have

$\int \frac{du}{(4u)^2}\ = \frac{1}{16} \int \frac{1}{u^2} du\ = \frac{1}{16} \int u^{-2} du\ = -\frac {1}{16u} + C = -\frac{1}{16(x^2-x)} + C$

The way he wrote it made it seem that $(4x^2-4x)^2=16(x^2-x)$ <-- NOT TRUE

Below are the correct calulations

$(4x^2-4x)^2=(4(x^2-x))^2=16(x^2-x)^2$ we used that middle one for our substitution $(4u)^2 = 16u^2$

Edit : Mistake fixed
• Feb 6th 2008, 12:40 AM
wingless
Quote:

Originally Posted by ThePerfectHacker
$(4x^2-4x)^2 = (4(x^2-x))^2 = 4^2(x^2-x) = 16(x^2-x)$

Isn't it $(4(x^2-x))^2 = 4^2(x^2-x)^2 = 16(x^2-x)^2$ ?

Quote:

Originally Posted by Lopared
Below are the correct calulations

$(4x^2-4x)^2=(4(x^2-x))^2=16(x^4-x^2)$

$(x^2 - x)^2 \neq x^4 - x^2$.
• Feb 6th 2008, 08:53 AM
Lopared
Quote:

Originally Posted by wingless
Isn't it $(4(x^2-x))^2 = 4^2(x^2-x)^2 = 16(x^2-x)^2$ ?

$(x^2 - x)^2 \neq x^4 - x^2$.

(Doh) yea your right. my mistake (fixed now)