# How do you integrate  (x^3)ln(x)dx?

Jul 27, 2016

${x}^{4} / 16 \left(4 \ln x - 1\right) + C$,

OR,

$\ln \left\{K {\left({x}^{4} / e\right)}^{{x}^{4} / 16}\right\}$

#### Explanation:

We use the Rule of Integration by Parts, which states,

$\int u v \mathrm{dx} = u \int v \mathrm{dx} - \int \left[\frac{\mathrm{du}}{\mathrm{dx}} \int v \mathrm{dx}\right] \mathrm{dx}$

We take u=lnx, v=x^3rArr(du)/dx=1/x,&, intvdx=x^4/4

Hence, $I = \int {x}^{3} \ln x \mathrm{dx} = {x}^{4} / 4 \cdot \ln x - \int \left(\frac{1}{x} \cdot {x}^{4} / 4\right) \mathrm{dx}$

$= \frac{{x}^{4} \ln x}{4} - \frac{1}{4} \int {x}^{3} \mathrm{dx} = \frac{{x}^{4} \ln x}{4} - {x}^{4} / 16$

$\therefore I = {x}^{4} / 16 \left(4 \ln x - 1\right) + C$

$I$ can further be shortened as under :-

$I = {x}^{4} / 16 \left(4 \ln x - 1\right) = {x}^{4} / 16 \left(\ln {x}^{4} - \ln e\right) = {x}^{4} / 16 \ln \left({x}^{4} / e\right)$

$\therefore I = \ln {\left({x}^{4} / e\right)}^{{x}^{4} / 16} + \ln K = \ln \left\{K {\left({x}^{4} / e\right)}^{{x}^{4} / 16}\right\}$

Enjoy Maths.!

Jul 27, 2016

$\frac{1}{4} {x}^{4} {\log}_{e} x - {x}^{4} / 16$

#### Explanation:

$\frac{d}{\mathrm{dx}} \left({x}^{n + 1} {\log}_{e} x\right) = \left(n + 1\right) {x}^{n} {\log}_{e} x + {x}^{n}$

so

$\int {x}^{n} {\log}_{e} x \mathrm{dx} = \frac{1}{n + 1} {x}^{n + 1} {\log}_{e} x - \int {x}^{n} / \left(n + 1\right) \mathrm{dx} + C$

Finally

$\int {x}^{n} {\log}_{e} x \mathrm{dx} = \frac{1}{n + 1} {x}^{n + 1} {\log}_{e} x - {x}^{n + 1} / {\left(n + 1\right)}^{2} + C$

for $n = 3$ we have the result

$\frac{1}{4} {x}^{4} {\log}_{e} x - {x}^{4} / 16 + C$