1. ## Integration by substitution

$\displaystyle Integrate: x root (4x-3)$ using substitution u = 4x-3

Sorry, don't know how to use the codes to make it look mathematical. Where can I learn to do that?

2. Originally Posted by nerdzor
$\displaystyle Integrate: x root (4x-3)$ using substitution u = 4x-3
do as they suggest

Let $\displaystyle u = 4x - 3 \implies \boxed{x = \frac 14(u + 3)}$

$\displaystyle \Rightarrow du = 4~dx$

$\displaystyle \Rightarrow \frac 14 ~du = dx$

By plugging in all these pieces, we obtain the integral,

$\displaystyle \frac 1{16} \int (u + 3) \sqrt u~du$

now continue, hopefully you see how we got to that

Sorry, don't know how to use the codes to make it look mathematical. Where can I learn to do that?
here

3. Originally Posted by Jhevon

$\displaystyle \int (u + 3) \sqrt u~du$
Am I able to expand $\displaystyle (u + 3) \sqrt u~du$ os that it becomes $\displaystyle u\sqrt u + 3\sqrt u$? Does it work like that?

4. Originally Posted by nerdzor
Am I able to expand $\displaystyle (u + 3) \sqrt u~du$ os that it becomes $\displaystyle u\sqrt u + 3\sqrt u$? Does it work like that?
Yes. And then write $\displaystyle u\sqrt u + 3\sqrt u = u^{3/2} + 3 u^{1/2}$