Evaluate the integral. $\int x^2 \log(4x) dx$

Question

The problem is $\int x^2 \log(4x) dx$

Here $\ln$ refers to the natural logarithm.

So far, I know $u = x^2$ and $du = 2x (dx)$.

So $dv = \ln(4x) dx$ and $v = 1/x$, but I don't know where to go from here.

Answer 1

Integrating by parts,

$$\int x^2\log 4xdx = \frac{1}{3}x^3\log4x-\int\frac{1}{3}x^3\cdot\frac{4}{4x}dx\\ =\frac{1}{3}x^3\log4x - \frac{1}{3}\int x^2dx\\ =\frac{1}{3}x^3\log4x - \frac{1}{9}x^3 + C$$

Answer 2

em,I think you have a little wrong in which $dv=ln(4x)dx$ imply to $v={\frac{1}{x}}$,

Hint : let $u=ln(4x)$ and $dv=(x^2)dx$

then $du=({\frac{1}{x}})dx$ and $v={\frac{(x^3)}{3}}$

thus we have, ${\frac{x^3}{3}}ln(4x)-\int{\frac{(x^2)}{3}}$, here it is trivial,I hope that it will be helpful.

Answer 3

$$\int x^2\ln (4x)dx = \dfrac {1}{3}\int \ln (4x)d(x^3)=\dfrac {1}{3}\left (x^3\ln (4x)-\int x^3d(\ln (4x))\right)$$ Now, $$\int x^3d(\ln (4x))=\int 4\dfrac {x^3}{4x} dx$$ Thus simplifying your problem considerably.