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

Dec 3, 2015

$- \frac{\ln \left({x}^{3}\right)}{3 {x}^{3}} - \frac{1}{3 {x}^{3}} + C = - \frac{\ln \left(x\right)}{{x}^{3}} - \frac{1}{3 {x}^{3}} + C$

#### Explanation:

Use integration-by-parts with $u = \ln \left({x}^{3}\right) = 3 \ln \left(x\right)$ and $\mathrm{dv} = \frac{1}{{x}^{4}} \mathrm{dx} = {x}^{- 4} \mathrm{dx}$ to get $\mathrm{du} = \frac{3}{x}$ and $v = {x}^{- 3} / \left(- 3\right)$.

Then, since $\int \setminus u \setminus \mathrm{dv} = u v - \int \setminus v \setminus \mathrm{du}$, we get

$\int \setminus \frac{\ln \left({x}^{3}\right)}{{x}^{4}} \setminus \mathrm{dx} = \int \setminus \frac{3 \ln \left(x\right)}{{x}^{4}} \setminus \mathrm{dx}$

$= - \frac{\ln \left({x}^{3}\right)}{3 {x}^{3}} - \int \setminus - {x}^{- 4} \setminus \mathrm{dx} = - \frac{\ln \left({x}^{3}\right)}{3 {x}^{3}} + \int \setminus {x}^{- 4} \setminus \mathrm{dx}$

$= - \frac{\ln \left({x}^{3}\right)}{3 {x}^{3}} + \frac{{x}^{- 3}}{- 3} + C$

$= - \frac{\ln \left({x}^{3}\right)}{3 {x}^{3}} - \frac{1}{3 {x}^{3}} + C = - \frac{\ln \left(x\right)}{{x}^{3}} - \frac{1}{3 {x}^{3}} + C$