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If the antiderivative of $\frac{1}{x^{2} \sqrt{1 + x^{2}}}$ is $- \frac{\sqrt{f (x)}}{x} + C$
then $f (x)$ is equal to

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1+1/x2

tan−1⁡x

1+x2

21+x2

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detailed solution

Correct option is C

Let I=∫dxx21+x2Put x=1/t, to be obtain I=−∫tdt1+t2=−1+t2+C=C−1+x2x Thus , f(x)=1+x2

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If the antiderivative of 1x21+x2 is −f(x)x+Cthen f(x) is equal to

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If the antiderivative of $\frac{1}{x^{2} \sqrt{1 + x^{2}}}$ is $- \frac{\sqrt{f (x)}}{x} + C$
then $f (x)$ is equal to