<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've got the right idea, but there are two places you erred:
$$ \int \frac{\frac{1}{2}\cdot4x-1+1}{\sqrt {4x-1}}\, \mathrm{d}x = \int \frac {2x - 1 + 1}{\sqrt{4x - 1}}\,dx \neq \frac{1}{2}\int \frac{4x-1+1}{\sqrt {4x-1}}\, \mathrm{d}x$$
You need to have: $$\int \dfrac{2x}{\sqrt{4x-1}}\,dx = \int \frac{\frac{1}2(4x-1) +\frac 12}{\sqrt {4x-1}}\, \mathrm{d}x = \frac 12\int \frac{(4x - 1) + 1}{\sqrt{4x-1}}$$
Here you also erred: $$ \frac{1}{2}\int\sqrt {u}+\frac{1}{\sqrt {u}}\cdot\frac{1}{4}\, \mathrm{d}u = \int \left(\frac 12 \sqrt u + \frac 1{8\sqrt u}\right)\,du \neq \frac{1}{8}\int \sqrt {u}+\frac{1}{\sqrt {u}}\, \mathrm{d}u $$
If you got the correct answer, it's because the second error reversed the first error.
If you go back and fix the first error, and work from there with the same idea (just be careful to note the scope of a a factor, and to distribute the factor, when needed!) you should obtain the correct result, and do so correctly.
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