Question 226112
<font face="Garamond" size="+2">


I'm going to presume you want to evaluate:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int{\frac{\ln(x)}{x^3}\,dx}]


Using integration by parts.  First split the integrand into two parts, *[tex \Large u] and *[tex \Large dv].  Select the parts so that when you differentiate *[tex \Large u] you get something simpler and when you integrate *[tex \Large dv] you get something simpler.  The split here should be *[tex \Large u\ =\ \ln(x)] because it differentiates to *[tex \Large \frac{1}{x}] and *[tex \Large dv\ =\ \frac{dx}{x^3}] because it integrates to *[tex \Large \frac{-1}{2x^2}]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u\ =\ \ln(x)]


therefore


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ du\ =\ \frac{1}{x}dx]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ dv\ =\ \frac{dx}{x^3}]


therefore


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ v\ =\ \int{\frac{dx}{x^3}}\ =\ \frac{-1}{2x^2}]


Now use the Integration by Parts Formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int{u\,dv}\ =\ uv\ -\ \int{v\,du}]


Substituting:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int{\frac{\ln(x)}{x^3}\,dx}\ =\ -\frac{\ln(x)}{2x^2}\ -\ \int{\left(\frac{-1}{2x^2}\right)\left(\frac{1}{x}\right)\,dx]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int{\left(\frac{-1}{2x^2}\right)\left(\frac{1}{x}\right)\,dx\ =\ \int{\frac{-1}{2x^3}\,dx\ =\ \frac{1}{4x^2}\ +\ C], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int{\frac{\ln(x)}{x^3}\,dx}\ =\ -\frac{\ln(x)}{2x^2}\ -\ \frac{1}{4x^2}\ +\ C]


Which still can be simplified somewhat, but I think you get the idea.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>