Question:
$\int \frac{\sqrt{1+x^{2}}}{x^{4}} d x$
Solution:
Let, $\mathrm{I}=\int \frac{\sqrt{1+x^{2}}}{x^{4}} d x$
$=\int \sqrt{\frac{1+x^{2}}{x^{2}}} \cdot \frac{1}{x^{3}} d x=\int \sqrt{\frac{1}{x^{2}}+1} \cdot \frac{1}{x^{3}} d x$
Taking, $\frac{1}{x^{2}}+1=t^{2}$
So, $\frac{-2}{x^{3}} d x=2 t d t \Rightarrow \frac{d x}{x^{3}}=-t d t$
$\therefore \mathrm{I}=\int t(t d t)=-\int t^{2} d t=-\frac{1}{3} t^{3}+\mathrm{C}$
Therefore, $I=-\frac{1}{3}\left(\frac{1}{x^{2}}+1\right)^{3 / 2}+C$
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