If the integral is
[tex]\displaystyle \int \frac{\ln(3x^2)}{x^5}\,\mathrm dx[/tex]
substitute u = ln(3x ²) and du = 6x/(3x ²) dx = 2/x dx.
Then x ² = exp(u)/3 and x ⁴ = exp(2u)/9.
The integral is transformed to
[tex]\displaystyle \int \frac{\ln(3x^2)}{x^5}\,\mathrm dx = \int \frac{\ln(3x^2)}{2x^4} \times \dfrac2x \,\mathrm dx \\\\ = \int \frac{u}{2\times\dfrac{e^{2u}}9}\,\mathrm du \\\\ = \frac92 \int ue^{-2u}\,\mathrm du[/tex]
Integrate by parts:
[tex]f = u \implies \mathrm df = \mathrm du \\\\ \mathrm dg = e^{-2u}\,\mathrm du \implies g = -\dfrac12 e^{-2u}[/tex]
[tex]\displaystyle \int ue^{-2u}\,\mathrm du = fg - \int g\,\mathrm df \\\\ = -\dfrac12 ue^{-2u} + \displaystyle \frac12 \int e^{-2u}\,\mathrm du \\\\ = -\frac12 ue^{-2u} - \frac14 e^{-2u} + C[/tex]
Then
[tex]\displaystyle \frac92 \int ue^{-2u}\,\mathrm du = -\frac94 ue^{-2u} - \frac98 e^{-2u} + C[/tex]
which in terms of x would be
[tex]\displaystyle \int \frac{\ln(3x^2)}{x^5}\,\mathrm dx = -\frac94\times\frac{\ln(3x^2)}{9x^4} - \frac98 \times \frac1{9x^4} + C \\\\ = \boxed{-\frac{\ln(3x^2)}{4x^4}-\frac1{8x^4}+C}[/tex]
