$\int \sqrt{\large\frac{1+x}{x^2}} . \{\log (1+ \large\frac{1}{x^2}) \} \times \large\frac{1}{x^3} $$dx$

$1+\large\frac{1}{x^2}$$=t$

$0- \large\frac{2}{x^3}$$dx=dt$

=> $\int \sqrt{t} \log (t) \times \large\frac{-dt}{2}$

=> $-\large\frac{-1}{2}$$ \int \sqrt t . \log t .dt$

=> $ \large\frac{-1}{2}$$ \{ \log.t \times t^{3/2} \times \large\frac{2}{3} -\int \large\frac{t^{3/2}}{t} \times$$ \frac{2}{3}dt \}$

=> $ \large\frac{-1}{2}$$ \{ \large\frac{2}{3} $$ (t)^{3/2} \log (t) - (t)^{\frac{2}{3}} \times (\large\frac{2}{3})^2+c\}$

$\large\frac{-1}{2} \bigg \{\frac{2}{3} (1+\frac{1}{x^2})^{3/2} $$\log (1+\frac{1}{x^2}) -(1+\frac{1}{x^2})^{3/2} \times \frac{4}{9}\bigg\}+c $

Hence a is the correct answer.