Answer
$\int \frac{(logx)^{2}}{x}dx=\frac{(lnx)^{3}}{3}+ Constant$
Work Step by Step
Evaluate $\int \frac{(logx)^{2}}{x}dx$.
Consider $ln x =t$ and $\frac{1}{x} dx = dt$
Thus, $\int \frac{(logx)^{2}}{x}dx=\int t^{2} dt$
$=\frac{t^{3}}{3}+ Constant$
$=\frac{(lnx)^{3}}{3}+ Constant$
Hence, $\int \frac{(logx)^{2}}{x}dx=\frac{(lnx)^{3}}{3}+ Constant$