<s>[INST] How to find the integral: $ \int \frac{2x}{\sqrt {4x-1}}\, \mathrm{d}x\;?$ My problem, is how to find the integral of $$ \int \frac{2x}{\sqrt {4x-1}}\, \mathrm{d}x$$
Can I do this?
$$ \int \frac{\frac{1}{2}\cdot4x-1+1}{\sqrt {4x-1}}\, \mathrm{d}x$$
$$ \frac{1}{2}\int \frac{4x-1+1}{\sqrt {4x-1}}\, \mathrm{d}x$$
$$ \frac{1}{2}\int \frac{4x-1}{\sqrt {4x-1}}+\frac{1}{\sqrt {4x-1}}\, \mathrm{d}x$$
$$ \frac{1}{2}\int \sqrt {4x-1}+\frac{1}{\sqrt {4x-1}}\, \mathrm{d}x$$
then
$$ 4x-1= u/'$$
$$ 4dx= du$$
$$ dx= \frac{1}{4}du$$
insert that in the integral
$$ \frac{1}{2}\int \sqrt {u}+\frac{1}{\sqrt {u}}\cdot\frac{1}{4}\, \mathrm{d}u$$
$$ \frac{1}{8}\int \sqrt {u}+\frac{1}{\sqrt {u}}\, \mathrm{d}u$$
and I get
$$\frac{1}{12}\cdot u\cdot \sqrt{u}+ \frac{1}{4}\cdot \sqrt{u}+C$$
$$\frac{1}{12}\cdot(4x-1)\cdot \sqrt{4x-1}+ \frac{1}{4}\cdot \sqrt{4x-1}+C$$
and that's the correct solution. But can I just rewrite $2x$ as $\frac{1}{2}\cdot4x-1+1$ and then put $\frac{1}{2}$ before the integral? Thank you in advance!
 [/INST] You can't do in that way, but you can salvage the idea:
$$
\int\frac{2x}{\sqrt{4x-1}}dx=\frac{1}{2}\int\frac{4x}{\sqrt{4x-1}}dx
=\frac{1}{2}\int\frac{4x-1+1}{\sqrt{4x-1}}dx
$$
and go on.
A possibly simpler strategy is doing the substitution
$$
\sqrt{4x-1}=2t
$$
where I use $2t$ because after squaring we get
$$
4x-1=4t^2
$$
so
$$
x=t^2+\frac{1}{4},\quad dx=2t\,dt
$$
and the integral becomes
$$
\int\frac{2t^2+\frac{1}{2}}{t}2t\,dt=\int(4t^2+1)\,dt=\frac{4}{3}t^3+t+c
$$
and the result will follow by substituting back $x$.
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