Per partes integration: $\int x^2 \ln x dx$

Question

Can I asked you guys for help with solving of this integral? Thank you.

It could go well with the per-parted method, but I am not able to finish it.

$$\int x^2 \ln x dx$$

Answer 1

Hint: Integrating by parts with $$u=lnx, dv=x^2dx$$

Answer 2

$$\int x^{2}\ln(x)\ dx=\left[\frac{1}{3}x^3\ln(x)\right]-\int \frac{1}{3}x^2dx=\frac{1}{3}x^3\ln(x) - \frac{1}{9}x^3 + C.$$

Answer 3

$$\int x^{2}\ln(x)dx=x^{2}(x\ln(x)-x)-\int 2x^{2}\ln(x)+2\int x^{2}dx$$

Hence,

$$3\int x^{2}\ln(x)dx=x^{2}(x\ln(x)-x)+\frac{2}{3}x^{3}+C$$

so

$$\int x^{2}\ln(x)dx=\frac{1}{3}x^{2}(x\ln(x)-x)+\frac{2}{9}x^{3}+D=\frac{1}{3}x^{3}\ln(x)-\frac{1}{9}x^{3}+D$$

Answer 4

Hint

Use $u=\log(x)$ and $v'=x^2~dx$. So $u'=\frac{dx}{x}$ and $v=\frac{x^3}{3}$

I am sure that you can take from here.