How do you find the indefinite integral of #int 1/(xlnx^3)#?

Question

Isabella Ackerman · Accepted Answer

Substitute #x=e^u# to get #1/3 ln(ln)x+C# #I=int1/(x ln x^3)dx=1/3 int 1/(x ln x)dx# by fundamental law of logarithms. Now substitute #x=e^u#, #ln x = u# and #dx=e^u du# to get #I=1/3int1/(e^u u)e^udu=1/3 int1/u du=1/3ln x + C#. (Or just substitute #x=e^u# at the start.)

Emily Ammerman · Answer

#int (dx)/(xlnx^3) = 1/3ln(ln x) +C#

We begin by exploiting the properties of logarithms and write:

#int (dx)/(xlnx^3) = int (dx)/(3xlnx)#

Now we note that #d(lnx) = (dx)/x#, so that:

#int (dx)/(3xlnx) = 1/3 int (d(lnx))/lnx= 1/3ln(ln x) +C#

Isaiah Allen · Answer

You may have also meant to type #int1/(x(lnx)^3)dx#. If this is the case, look here; if not, you can still learn something from this! If we do indeed have #int1/(x(lnx)^3)dx#, let #u=lnx#. This implies that #du=1/xdx#. Then we have the equivalent integrals #int1/(lnx)^3(1/xdx)=int1/u^3du=intu^-3du#. Now we can use #intu^ndu=u^(n+1)/(n+1)+C#, where #n!=-1#, which is not an issue here. The integral then becomes #(lnx)^-2/(-2)+C#, or #(-1)/(2(lnx)^2)+C#.

Paisley Johnson · Answer

undefined To find the indefinite integral of $\int \frac{1}{x \ln(x^3)}$, you can perform a substitution. Let $u = \ln(x^3)$, which implies $x^3 = e^u$ and $x = e^{u/3}$. Then, find $du$ and substitute $u$ and $du$ into the integral. This will transform the integral into a simpler form, which you can then integrate with respect to $u$. After integrating with respect to $u$, substitute $u = \ln(x^3)$ back in to obtain the final result.