Question

Solve $\int {\left(\frac{\sqrt{x}+1}{x}\right)}^2$ $dx$, $xϵ(0,\infty )$

Answer 1

$\int (\frac{\sqrt{x}+1}{x}{)}^{2}dx$

$=\int \frac{(\sqrt{x}+1{)}^2}{x^2}dx=\int \frac{x+1+2\sqrt{x}}{x^2}dx$

$=\int \frac{1}{x}dx+\int x^{-2}dx+2\int x^{-\frac{3}{2}}dx$

$=logx+\frac{x^{-1}}{-1}+2\frac{x^{-\frac{1}{2}}}{-\frac{1}{2}}$

$=logx-\frac{1}{x}-\frac{4}{\sqrt{x}}$