Question

Evaluate $\int {\left(\frac{1+logx}{x}\right)}^{2}dx.$

Answer 1

$\int {\left(\frac{1+logx}{x}\right)}^{2}dx$

Let $y=1+logx\Rightarrow e^{y-1}=x$

$\frac{dy}{dx}=\frac{1}{x}$

$\int \frac{y^2}{\left(e^{y-1}\right)}dy$

$=\int e^{1-y}{y}^{2}dy$

$=e\int e^{-y}{y}^{2}dy$

$=e[y^2\frac{e^{-y}}{(-1)}-\int 2y\frac{e^{-y}}{(-1)}dy]$

$=e[-y^2{e}^{-y}+2[y\frac{e^{-y}}{(-1)}-\frac{e^{-y}}{(-1{)}^2}]]+C$

$=e[-y^2{e}^{-y}-2y{e}^{-y}-2e^{-y}]+C$

$=e^{1-y}[-y^2-2y-2]+C$

$\therefore \int {\left(\frac{1+logx}{x}\right)}^{2}dx=e^{1-y}[-y^2-2y-2]+C$