Answer
$ - \frac{2}{3}{\left( {1 + \frac{1}{x}} \right)^{3/2}} + C$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{{x^2}}}\sqrt {1 + \frac{1}{x}} } dx \cr
& {\text{Let }}u = 1 + \frac{1}{x} \to du = - \frac{1}{{{x^2}}}dx,{\text{ }} - du = \frac{1}{{{x^2}}}dx \cr
& {\text{Applying the substitution}}{\text{, we obtain}} \cr
& \int {\frac{1}{{{x^2}}}\sqrt {1 + \frac{1}{x}} } dx = \int {\overbrace {\sqrt {1 + \frac{1}{x}} }^{\sqrt u }} \overbrace {\frac{1}{{{x^2}}}dx}^{ - du} \cr
& = \int {\sqrt u \left( { - du} \right)} \cr
& = - \int {{u^{1/2}}du} \cr
& {\text{Integrate by the power rule}} \cr
& = - \frac{{{u^{1/2 + 1}}}}{{1/2 + 1}} + C \cr
& = - \frac{{{u^{3/2}}}}{{3/2}} + C \cr
& = - \frac{{2{u^{3/2}}}}{3} + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}u = 1 + \frac{1}{x} \cr
& = - \frac{2}{3}{\left( {1 + \frac{1}{x}} \right)^{3/2}} + C \cr} $$
