First slide

Methods of integration

Question

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

Easy

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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