What is #int1/(xln^3x)dx# ?

#int1/(xln^3x)dx#

1 Answer
Apr 20, 2018

Answer:

#int 1/(xln^3x)dx =-1/(2ln^2x) + C#

Explanation:

We will use integration by substitution:

Let #u = lnx#. Then #(du)/dx = 1/x => dx=xdu#.

Thus,

#int 1/(xln^3x) dx = int 1/(xu^3) * xdu = int 1/u^3 du#

As you can see, the #x# cancells nicely.

We have to apply the power rule now:

#int u^n du = u^(n+1)/(n+1) + C# with #n = -3#.

#int 1/u^3 du = u^(-2)/-2 +C= -1/(2u^2)+C#

Undo the substitution :

# = -1/(2ln^2x) + C#

#color(red)( :. int 1/(xln^3x) dx =-1/(2ln^2x) + C#