First slide

Methods of integration

Question

$\int \frac{2^{y}}{\sqrt{4^{y} + 36}} d y is equal to$

Moderate

A

$\log |2^{y} + \sqrt{4^{y} + 36}| + k$
B

$\frac{- 1}{\log 2} \log |2^{y} + \sqrt{4^{y} + 36}| + k$
C

$\frac{1}{\log 2} |2^{y} + \sqrt{4^{y} + 36}| + k$
D

$\frac{1}{\log 2} \log |2^{y} + \sqrt{2^{y} + 36}| + k$

Solution

$\int \frac{2^{y}}{\sqrt{4^{y} + 36}} d y = \int \frac{2^{y} d y}{\sqrt{{(2^{y})}^{2} + 36}}$
$Put 2^{y} = t$
$\begin{array}{l} 2^{y} \log 2 d y = d t \\ = \frac{1}{\log 2} \int \frac{d t}{\sqrt{t^{2} + 6^{2}}} \\ = \frac{1}{\log 2} \log |t + \sqrt{t^{2} + 6^{2}}| + k \\ = \frac{1}{\log 2} \log |2^{y} + \sqrt{4^{y} + 36}| + k \end{array}$

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