Answer
$$\frac{1}{2}\tan^2 (\ln t) -\ln |\sec (\ln t) |+c$$
Work Step by Step
Given
$$\int \frac{\tan^3 \ln t}{t}dt $$
Let $$ u =\ln t \ \ \ \to \ \ \ du=\frac{dt}{t}$$
Then
\begin{align*}
\int \tan^3 udu&= \int \tan^2 u \tan udu\\
&= \int (\sec^2 u-1) \tan udu\\
&=\frac{1}{2}\tan^2 u -\ln |\sec u |+c\\
&=\frac{1}{2}\tan^2 (\ln t) -\ln |\sec (\ln t) |+c
\end{align*}
