I don't speak Romanian, but the closest translation for this suggests you're trying to compute
[tex]\displaystyle \int x^3 \ln(x)^2 \, dx[/tex]
Integrate by parts:
[tex]\displaystyle \int x^3 \ln(x)^2 \, dx = uv - \int v \, du[/tex]
where
u = ln(x)² ⇒ du = 2 ln(x)/x dx
dv = x³ dx ⇒ v = 1/4 x⁴
[tex]\implies \displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac12 \int x^3 \ln(x) \, dx[/tex]
Integrate by parts again:
[tex]\displaystyle \int x^3 \ln(x) \, dx = u'v' - \int v' du'[/tex]
where
u' = ln(x) ⇒ du' = dx/x
dv' = x³ dx ⇒ v' = 1/4 x⁴
[tex]\implies \displaystyle \int x^3 \ln(x) \, dx = \frac14 x^4 \ln(x) - \frac14 \int x^3 \, dx[/tex]
So, we have
[tex]\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac12 \left(\frac14 x^4 \ln(x) - \frac14 \int x^3 \, dx \right)[/tex]
[tex]\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac18 \int x^3 \, dx[/tex]
[tex]\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac18 \left(\frac14 x^4\right) + C[/tex]
[tex]\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac1{32} x^4 + C[/tex]
[tex]\boxed{\displaystyle \int x^3 \ln(x)^2 \, dx = \frac1{32} x^4 \left(8\ln(x)^2 - 4\ln(x) + 1\right) + C}[/tex]
