Evaluate the integral. ∫012ln⁡(x)xdx

Step 1
Consider the provided integral,
${\displaystyle {\int }_0^1\frac{2\ln \left(x\right)}{\sqrt{x}}dx}$
Evaluate the above integral.
Step 2
Now, solve the above integral using by part formula.
${\displaystyle {\int }_0^1\frac{2\ln \left(x\right)}{\sqrt{x}}dx=2\cdot {\int }_0^1\frac{\ln \left(x\right)}{\sqrt{x}}dx}$
${\displaystyle =2{[2\sqrt{x}\ln \left(x\right)-\int \frac{2}{\sqrt{x}}dx]}_0^1}$
${\displaystyle =2{[2\sqrt{x}\ln \left(x\right)-4\sqrt{x}]}_0^1}$
=2(-4)
=-8

${\displaystyle \int \frac{2\ln \left(x\right)}{\sqrt{x}}dx}$
Lets

The integration process can be simplified by making a change of variables:
$t=\sqrt{x},x=t^2,dx=2\ast t\ast dt$
Then the original integral can be written as follows: The
$\int 4\ast \ln (t^2)\ast dt$
$\int 4\ast \ln (x^2)\ast dx$
formula for integration by parts:
$\int U(x)\ast dV(x)=U(x)\ast V(x)-\int V(x)\ast dU(x)$
Put
$U=4\ln (x^2)$
dV=dx
Then:
$dU=\frac{8}{x}\ast dx$
V=x
Therefore:
$\int 4\ast \ln (x^2)\ast dx=4\ast x\ast \ln (x^2)-\int 8\ast dx$
Find the integral
$\int 8\ast dx=8\ast x$
Answer:
$\int 4\ast \ln (x^2)=4\ast x\ast \ln (x^2)-8\ast x+C$
or
$\int 4\ast \ln (x^2)=4\ast x\ast (\ln (x^2)-2)+C$
To write the final answer, it remains to substitute sqrt (x) instead of t.
$4\sqrt{x}(\ln (x)-2)+C$
Let's calculate a definite integral:
${\int }_0^1\frac{2\ast \ln (x)}{\sqrt{x}}\ast dx=(4\sqrt{x}\ast (\ln (x)-2)){|}_0^1$
F(1)=-8
F(0)=
I=-8-(0)=-8