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# What is the integral of "ln(x^3)/x dx"?

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- PatrickLv 68 years agoFavorite Answer
Recall integration by parts: ∫udv = uv - ∫vdu.

∫(ln(x^3)/x)dx

Let u = ln(x^3), du = ((3x^2)/(x^3))dx = (3/x)dx.

Then dv = dx/x, v = ln|x|.

So ∫(ln(x^3)/x)dx = ln|x|•ln(x^3) - 3∫(ln|x|/x)dx.

Let w = ln|x|, dw = dx/x.

Then ln|x|•ln(x^3) - 3∫(ln|x|/x)dx

= ln|x|•ln(x^3) - 3∫wdw

= ln|x|•ln(x^3) - (3w^2)/2 + C

= ln|x|•ln(x^3) - (3•ln|x|^2)/2 + C.

As x > 0 is required from the get go [ln(x^3)/x is only real for x > 0],

we can just say the integral is ln(x)•ln(x^3) - (3•ln(x)^2)/2 + C.

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