Answer
$\dfrac{1}{8}(\ln 4)^4 +c$
Work Step by Step
Solve $\int_1^4\dfrac{(\ln x)^3}{2x}dx$
Let us $p= \ln x$ and $dp=\dfrac{1}{x}dx$
This implies $\int_1^4\dfrac{(\ln x)^3}{2x}dx= \dfrac{1}{2} \int_1^4 u^3 du$
$= \dfrac{1}{8} [u^4]_1^4+c$
$= \dfrac{1}{8}(\ln 4)^4 +c$
Hence, $\int_1^4\dfrac{(\ln x)^3}{2x}dx=\dfrac{1}{8}(\ln 4)^4 +c$
