1. ## Integration by substitution

$\displaystyle \int \frac{2x-1}{(4x^2-4x)^{2}} dx\$

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

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

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

but obviously thats not it.

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

In this case what would u be? that the integral would be
.
$\displaystyle (4x^2-4x)^2 = 16(x^2 - x)$.
Let $\displaystyle u=x^2-x$.

3. Originally Posted by ThePerfectHacker
$\displaystyle (4x^2-4x)^2 = 16(x^2 - x)$.
Let $\displaystyle u=x^2-x$.
....dumb question ... but how did u get that? the 16(x^2-x)??

4. Originally Posted by simsima_1
....dumb question ... but how did u get that? the 16(x^2-x)??
$\displaystyle (4x^2-4x)^2 = (4(x^2-x))^2 = 4^2(x^2-x)^2 = 16(x^2-x)^2$

EDIT: Mistake fixed.

5. $\displaystyle (4x^2-4x)^2=(4(x^2-x))^2$
if $\displaystyle x^2-x = u$ then $\displaystyle (2x-1)dx=du$ so you have

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

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

Below are the correct calulations

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

Edit : Mistake fixed

6. Originally Posted by ThePerfectHacker
$\displaystyle (4x^2-4x)^2 = (4(x^2-x))^2 = 4^2(x^2-x) = 16(x^2-x)$
Isn't it $\displaystyle (4(x^2-x))^2 = 4^2(x^2-x)^2 = 16(x^2-x)^2$ ?

Originally Posted by Lopared
Below are the correct calulations

$\displaystyle (4x^2-4x)^2=(4(x^2-x))^2=16(x^4-x^2)$
$\displaystyle (x^2 - x)^2 \neq x^4 - x^2$.

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

$\displaystyle (x^2 - x)^2 \neq x^4 - x^2$.
yea your right. my mistake (fixed now)