If the antiderivative of 1×21+x2 is −f(x)x+Cthen f(x) is equal to

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

A
$1 + 1 / x^{2}$
B
$\tan^{- 1} x$
C
$1 + x^{2}$
D
$2 \sqrt{1 + x^{2}}$

Solution:

Let $I = \int \frac{d x}{x^{2} \sqrt{1 + x^{2}}}$

Put $x = 1 / t$ , to be obtain

$\begin{array}{l} I = - \int \frac{t d t}{\sqrt{1 + t^{2}}} = - \sqrt{1 + t^{2}} + C \\ = C - \frac{\sqrt{1 + x^{2}}}{x} \end{array}$

Thus , $f (x) = 1 + x^{2}$

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