∫11+2xdx=1log⁡2log⁡f(x)+C then f(0)=

I=∫11+2xdx=∫2x2x1+2xdx Let 2x+1=t2⇒2xlog⁡2dx=2tdt2x=t2−1 2xdx=2tlog⁡2dtI=∫1t2−1t22log⁡2tdtI=2log⁡2∫1t2−1dt=2log⁡212log⁡t−1t+1+C=1log⁡2log⁡1+2x−11+2x+1+C

∫11+2xdx=1log⁡2log⁡f(x)+C then f(0)=

I=∫11+2xdx=∫2x2x1+2xdx Let 2x+1=t2⇒2xlog⁡2dx=2tdt2x=t2−1 2xdx=2tlog⁡2dtI=∫1t2−1t22log⁡2tdtI=2log⁡2∫1t2−1dt=2log⁡212log⁡t−1t+1+C=1log⁡2log⁡1+2x−11+2x+1+C

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$\int \frac{1}{\sqrt{1 + 2^{x}}} d x = \frac{1}{\log 2} \log f (x) + C then f (0) =$

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$2 \sqrt{2} + 3$

$3 - 2 \sqrt{2}$

None

answer is A.

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Detailed Solution

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

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