I=∫2x4x+2x+1+2dx=∫2x2x2+22x1+1+1dx Let 1+2x=t→2xlog⁡2dx=dt2xdx=1log⁡2dtI=1log⁡2∫1t2+1dt=1log⁡2tan−1⁡(t)+C=tan−1⁡2x+1log⁡2+C

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$\int \frac{2^{x}}{4^{x} + 2^{x + 1} + 2} d x =$

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$\frac{\tan^{- 1} (2^{x})}{\log 2} + C$

$\log 2 (2^{x} + 1) + C$

$\frac{\tan^{- 1} (2^{x} + 1)}{\log 2} + C$

None

answer is C.

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Detailed Solution

$\begin{array}{l} I = \int \frac{2^{x}}{4^{x} + 2^{x + 1} + 2} d x \\ = \int \frac{2^{x}}{{(2^{x})}^{2} + 2 (2^{x}) 1 + 1 + 1} d x \\ Let 1 + 2^{x} = t \to 2^{x} \log 2 d x = d t \\ 2^{x} d x = \frac{1}{\log 2} d t \\ I = \frac{1}{\log 2} \int \frac{1}{t^{2} + 1} d t \\ = \frac{1}{\log 2} \tan^{- 1} (t) + C \\ = \frac{\tan^{- 1} (2^{x} + 1)}{\log 2} + C \end{array}$