Adrianna Macias

2022-07-19

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

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

Here $\mathrm{ln}$ refers to the natural logarithm.

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

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

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

Here $\mathrm{ln}$ refers to the natural logarithm.

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

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

eishale2n

Beginner2022-07-20Added 15 answers

Integrating by parts,

$\int {x}^{2}\mathrm{log}4xdx=\frac{1}{3}{x}^{3}\mathrm{log}4x-\int \frac{1}{3}{x}^{3}\cdot \frac{4}{4x}dx\phantom{\rule{0ex}{0ex}}=\frac{1}{3}{x}^{3}\mathrm{log}4x-\frac{1}{3}\int {x}^{2}dx\phantom{\rule{0ex}{0ex}}=\frac{1}{3}{x}^{3}\mathrm{log}4x-\frac{1}{9}{x}^{3}+C$

$\int {x}^{2}\mathrm{log}4xdx=\frac{1}{3}{x}^{3}\mathrm{log}4x-\int \frac{1}{3}{x}^{3}\cdot \frac{4}{4x}dx\phantom{\rule{0ex}{0ex}}=\frac{1}{3}{x}^{3}\mathrm{log}4x-\frac{1}{3}\int {x}^{2}dx\phantom{\rule{0ex}{0ex}}=\frac{1}{3}{x}^{3}\mathrm{log}4x-\frac{1}{9}{x}^{3}+C$

Levi Rasmussen

Beginner2022-07-21Added 6 answers

$\int {x}^{2}\mathrm{ln}(4x)dx={\displaystyle \frac{1}{3}}\int \mathrm{ln}(4x)d({x}^{3})={\displaystyle \frac{1}{3}}({x}^{3}\mathrm{ln}(4x)-\int {x}^{3}d(\mathrm{ln}(4x)))$

$\int {x}^{3}d(\mathrm{ln}(4x))=\int 4{\displaystyle \frac{{x}^{3}}{4x}}dx$

Thus simplifying your problem considerably.

$\int {x}^{3}d(\mathrm{ln}(4x))=\int 4{\displaystyle \frac{{x}^{3}}{4x}}dx$

Thus simplifying your problem considerably.

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