What is #int lnx^2 / lnx^3#?

1 Answer
Nov 14, 2015

Answer:

#intln(x^2)/ln(x^3)dx = (2x)/3 + c#

Explanation:

Well, when integrating you should always have #d[arg]#, where [arg] is a variable, it's like #int# is "(" and that is the ")". It's more important because it tell us with which variable you want us to integrate.

In this case it's pretty clear that's #x# but as a general rule it's important to specify.

Anyhow, using the property of logarithms we have

#intln(x^2)/ln(x^3)dx = int(2ln(x))/(3ln(x))dx = int(2dx)/3 = (2x)/3 + c#

Assuming # x in RR, x > 0, x != 1# so as to avoid complex numbers when doing those algebrisms and/or having the function defined in the reals at those points.