What is the technique to integrate $ \int_{0}^{3} x\ln(x) dx ?$ What is the technique to integrate $$ \int_{0}^{3} x\ln(x) dx  ?$$
Why did one textbook evaluate this integral using the integration technique for improper integrals with a discontinuity on its integrand? Isn't it that we can only use that said technique if the integrand has an infinite discontinuity at some number on the interval of integration and if $$ \lim_{x\to 0+} \vert x\ln(x) \rvert = +\infty ?$$
But our integrand has no infinity discontinuity, but a removable disconituity at $x=0$. Also, $ \lim_{x\to 0+} \vert \ln(x) \rvert \neq +\infty $
I am thinking of redefining the integrand first before integrating. But is this even possible?

Hint:
As outlined in my comments

*

*Integration by parts gives
$$
\int x\log(x)\,dx =\frac{1}{2}\Big(\int \log(x)\,dx^2\Big)=\frac12\Big(x^2\log(x)-\int x\,dx\Big)=\frac{1}{2}x^2\log(x)-\frac{1}{4}x^2 +C
$$


*The main issue you seemed to have is the value of $\lim_{x\rightarrow0+}x\log(x)$. This limit exists and is $0$. To convince yourself of this, you may use L'Hospital rule:
$$
x\log(x)=\frac{\log(x)}{1/x}\sim\frac{1/x}{-1/x^2}=-x\xrightarrow{x\rightarrow0+}0
$$
There are other ways to do this but that will do.


*Putting things together, you have that the integral you are interested to evaluate is
$$\lim_{a\rightarrow0+}\Big(\frac{1}{2}x^2\log(x)-\frac{1}{4}x^2\Big)\Big|^3_a=\frac{9}{2}\log 3-\frac{9}{4}-\lim_{a\rightarrow0+}\Big(\frac{1}{2}a^2\log(a)-\frac{a^2}{4}\Big)=?
$$
I hope this helps you and you can estimate the last bit.
