integrate: lnx/x^3

[Melissaherman]

Hi, I'm doing improper integrals and once I figure out the integration method I'll be able to finish the problem.

The integral is lnx/x^3
And I dont even know where to start. I was thinking a substitution of lnx, giving me 1/x du, but I'm not sure how to manipulate that to give me x^3 in the denom.

Thanks for any suggestions!

 

[pka]

\(\displaystyle \L y' = \frac{{\ln (x)}}{{x^2 }}\quad \Rightarrow \quad y = - \frac{{\ln (x)}}{{2x^2 }} - \frac{1}{{4x^2 }}.\)

Do the derivative. You will see how it works.
The process requires "parts".

 

[soroban]

Hello, Melissaherman!

You had the right approach.
. . What stopped you?

\(\displaystyle \L I \:=\:\int\frac{\ln x}{x^3}\,dx\)

By parts

\(\displaystyle \begin{array}{ccccccc}u & = & \ln x & \quad & dv & = & x^{-3}dx \\ \\ \\
du & = & \frac{dx}{x} & \quad & v & = & -\frac{1}{2}x^{-2} \end{array}\)

We have: \(\displaystyle \L\:I \:=\:-\frac{1}{2}x^{-2}\cdot\ln x\,-\,\int\left(-\frac{1}{2}x^{-2}\right)\left(\frac{dx}{x}\right)\)

. . . . . . . .\(\displaystyle \L I\;=\;-\frac{1}{2}x^{-2}\cdot\ln x\,+\,\frac{1}{2}\int x^{-3}\,dx\)

. . . . . . . .\(\displaystyle \L I\;=\;-\frac{1}{2}x^{-2}\cdot\ln x\,+\,\frac{1}{2}\left(-\frac{1}{2}x^{-2}\right) \,+\,C\)

. . . . . . . .\(\displaystyle \L I\;=\;-\frac{1}{2}x^{-2}\cdot\ln x \,- \,\frac{1}{4}x^{-2}\,+\,C\)

. . . . . . . .\(\displaystyle \L I \;=\;-\frac{1}{4x^2}\left(2\cdot\ln x\,+\,1\right)\,+\,C\)

 

[Count Iblis]

You can also write down the integral of x^(-p) and then differentiate w.r.t. p. That's much faster and easier than partial integration. In fact, most "partial integration" problems can usually be done much faster by differentiation w.r.t. a suitably chosen parameter.