What is $int(lnx^3)/x^4 dx$ ?

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Answer 1 · 2015-12-03T03:54:20

$-(ln(x^3))/(3x^3)-1/(3x^3)+C=-(ln(x))/(x^3)-1/(3x^3)+C$

Use integration-by-parts with $u=ln(x^3)=3ln(x)$ and $dv=1/(x^4)dx=x^(-4)dx$ to get $du=3/x$ and $v=x^(-3)/(-3)$ .

Then, since $int\ u\ dv=uv-int\ v\ du$ , we get

$int\ (ln(x^3))/(x^4)\ dx= int\ (3ln(x))/(x^4)\ dx$

$=-(ln(x^3))/(3x^3)-int\ -x^(-4)\ dx=-(ln(x^3))/(3x^3)+int\ x^(-4)\ dx$

$=-(ln(x^3))/(3x^3)+(x^(-3))/(-3)+C$

$=-(ln(x^3))/(3x^3)-1/(3x^3)+C=-(ln(x))/(x^3)-1/(3x^3)+C$

What is int(lnx^3)/x^4 dxint(lnx^3)/x^4 dx?