Question

The integral of In3x²/×⁵

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Answer 1

If the integral is

`\displaystyle \int (\ln(3x^2))/(x^5)\,\mathrm dx`

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

`\displaystyle \int (\ln(3x^2))/(x^5)\,\mathrm dx = \int (\ln(3x^2))/(2x^4) * \frac2x \,\mathrm dx \\\\ = \int \frac{u}{2*\frac{e^(2u)}9}\,\mathrm du \\\\ = \frac92 \int ue^(-2u)\,\mathrm du`

Integrate by parts:

`f = u \implies \mathrm df = \mathrm du \\\\ \mathrm dg = e^(-2u)\,\mathrm du \implies g = -\frac12 e^(-2u)`

`\displaystyle \int ue^(-2u)\,\mathrm du = fg - \int g\,\mathrm df \\\\ = -\frac12 ue^(-2u) + \displaystyle \frac12 \int e^(-2u)\,\mathrm du \\\\ = -\frac12 ue^(-2u) - \frac14 e^(-2u) + C`

Then

`\displaystyle \frac92 \int ue^(-2u)\,\mathrm du = -\frac94 ue^(-2u) - \frac98 e^(-2u) + C`

which in terms of x would be

`\displaystyle \int (\ln(3x^2))/(x^5)\,\mathrm dx = -\frac94*(\ln(3x^2))/(9x^4) - \frac98 * \frac1{9x^4} + C \\\\ = \boxed{-(\ln(3x^2))/(4x^4)-\frac1{8x^4}+C}`