First slide

Methods of integration

Question

If $\int \frac{\log (x + \sqrt{1 + x^{2}})}{\sqrt{1 + x^{2}}} d x = g \circ f (x) +$
Const. then

Moderate

A

$f (x) = \log (x + \sqrt{x^{2} + 1})$
B

$f (x) = \log (x + \sqrt{x^{2} + 1})$ and $g (x) = x^{2}$
C

$f (x) = \log (x + \sqrt{x^{2} + 1}) and g (x) = x^{2} / 2$
D

$f (x) = x^{2} / 2 and g (x) = \log (x + \sqrt{x^{2} + 1})$

Solution

Putting $\log (x + \sqrt{1 + x^{2}}) = t$ , we have
$\frac{1}{x + \sqrt{1 + x^{2}}} \times (1 + \frac{x}{\sqrt{1 + x^{2}}}) d x = d t i.e. \frac{d x}{\sqrt{1 + x^{2}}} = d t$
So, $\int \frac{\log (x + \sqrt{1 + x^{2}})}{\sqrt{1 + x^{2}}} d x = \int t d t = \frac{1}{2} t^{2} + C$
$= \frac{1}{2} {(\log (x + \sqrt{x^{2} + 1}))}^{2} + C$
Thus $f (x) = (\log (x + \sqrt{x^{2} + 1}))$ and $g (x) = x^{2} / 2$ .

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