First slide

Methods of integration

Question

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

Moderate

A

$\log |a^{x} + \sqrt{{(a^{x})}^{2} + 2}| + C$
B

$\log |a^{x} + \sqrt{{(a^{x})}^{2} - 2}| + C$
C

$\frac{1}{\log a} \log |a^{x} + \sqrt{{(a^{x})}^{2} + 1}| + C$
D

$\frac{1}{\log a} \log |a^{x} + \sqrt{{(a^{x})}^{2} - 1}| + C$

Solution

$Let I = \int \frac{a^{\frac{x}{2}} d x}{\sqrt{\frac{{(a^{x})}^{2} + 1}{a^{x}}}}$
$I = \int \frac{a^{\frac{x}{2}} a^{\frac{x}{2}} d x}{\sqrt{{(a^{x})}^{2} + 1}} = \int \frac{a^{x} d x}{\sqrt{{(a^{x})}^{2} + 1}}$
$Put a^{x} = t \Rightarrow a^{x} d x = \frac{1}{\log a} d t$
$\begin{array}{l} I = \int \frac{d t}{\log a} \frac{1}{\sqrt{(t^{2} + 1)}} = \frac{1}{\log a} \log |t + \sqrt{t^{2} + 1}| + C \\ = \frac{1}{\log a} \log |a^{x} + \sqrt{{(a^{x})}^{2} + 1}| + C \end{array}$

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