Q:

Evaluating $\int \sqrt{x}\ln(2x)\:\mathrm{d}x$

I am learning to an exam and stumbled upon this integral.
$$\int \sqrt{x}\ln(2x)\:\mathrm{d}x$$
the result should be $\frac{2}{3}\sqrt{x^{3}}\left(\ln(2x)-\frac{2}{3}\right)+C$, but I am still unable to get there… Every help will be appreciated.

A:

Hint
Integrate by parts with $u=\log(2x)$ and $v'=\sqrt x ~dx$. So $u'=\frac{1}{x}~dx$ and $v=\frac{2 x^{3/2}}{3}$.
I am sure that you can take from here.

A:

I think I'd start this by substituting $x=y^2$
$$\int 2y^2\ln2y^2dy$$
Now use properties of logarithms.
$$\int2y^2\ln2dy+\int4y^2\ln ydy$$
$$u=\ln y,du=\frac{dy}y$$
$$dv=4y^2dy,v=\frac43y^3$$
$$\frac23y^3\ln2+\frac43y^3\ln y-\int\frac43y^2dy=$$
$$\frac23y^3\ln2+\frac23y^3\ln y^2-\frac49y^3+C=$$
$$\frac23y^3\ln2y^2-\frac49y^3+C$$
Just a little rearranging from here and substituting back for $x$ should get you the right form.

