How do you integrate (lnx)2x3?

1 Answer
Douglas K.
Jun 2, 2017

Begin with integration by parts:

∫udv=uv−∫vdu

let u=(ln(x))2 and dv=x−3dx

then du=2ln(x)xdx and v=−12x−2

∫(ln(x))2x3dx=−12(ln(x)x)2−∫(−12x2)(2ln(x)x)dx

∫(ln(x))2x3dx=−12(ln(x)x)2+∫ln(x)x3dx

Integrate by parts:

∫udv=uv−∫vdu

let u=ln(x) and dv=x−3dx

then du=1xdx and v=−12x−2

∫(ln(x))2x3dx=−12(ln(x)x)2−ln(x)2x2−∫(−12x−2)(1x)dx

∫(ln(x))2x3dx=−12(ln(x)x)2−ln(x)2x2+12∫1x3dx

We have already done the last integral:

∫(ln(x))2x3dx=−12(ln(x)x)2−ln(x)2x2−14x2+C

Simplify over a common denominator:

∫(ln(x))2x3dx=−2(ln(x))2+2ln(x)+14x2+C

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