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$\int \frac{1}{x^{2} \sqrt{1 + x^{2}}} d x is equal to$

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−1+x2x+C

1+x2x+C

−x1+x2+C

x1+x2+C

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detailed solution

Correct option is A

Let I=∫1x21+x2dx=∫1x2x1+1x2dx Put 1+1x2=t⇒dxx3=−12dtI=∫−12tdt=−t+C=−1+1x2+C=−x2+1x+C

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∫1x21+x2dx is equal to

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$\int \frac{1}{x^{2} \sqrt{1 + x^{2}}} d x is equal to$