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Definite Integration — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration5 Marks
Question
Evaluate : $\int_0^2 \frac{2^x}{2^x\left(1+4^x\right)} \cdot d x$
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Answer

Let $I=\int_0^2 \frac{2^x}{2^x\left(1+4^x\right)} \cdot d x$
put $2^x=t \quad \therefore \quad 2^x \cdot \log 2 \cdot d x=1 \cdot d t$
As $x$ varies from 0 to $2, t$ varies from 1 to 4
$
\begin{aligned}
& =\int_1^4 \frac{\frac{1}{\log 2}}{t\left(1+t^2\right)} \cdot d t \\
& =\frac{1}{\log 2} \cdot \int_1^4 \frac{1}{t\left(1+t^2\right)} \cdot d t \\
& =\frac{1}{\log 2} \cdot \int_1^4 \frac{1+t^2-t^2}{t\left(1+t^2\right)} \cdot d t
\end{aligned}
$
may be solved by method of partial fraction
$
\begin{aligned}
& =\frac{1}{\log 2} \cdot \int_1^4\left[\frac{1+t^2}{t\left(1+t^2\right)}-\frac{t^2}{t\left(1+t^2\right)}\right] \cdot d t \\
& =\frac{1}{\log 2} \cdot \int_1^4\left[\frac{1}{t}-\frac{t}{1+t^2}\right] \cdot d t \\
& =\frac{1}{\log 2} \cdot\left[\int_1^4 \frac{1}{t} \cdot d t-\frac{1}{2} \int_1^4 \frac{2 t}{1+t^2} \cdot d t\right]
\end{aligned}
$
$\begin{aligned} & =\frac{1}{\log 2} \cdot\left[\log (t)-\frac{1}{2} \log \left(1+t^2\right)\right]_1^4 \\ & =\frac{1}{\log 2} \cdot\left[\left(\log 4-\frac{1}{2} \log 17\right)-\right. \\ & =\frac{1}{\log 2} \cdot\left[\log 4-\frac{1}{2} \log 17+\frac{1}{2} \log 2\right] \\ & \because \log 1=0 \\ & =\frac{1}{\log 2} \cdot\left[\log \frac{4 \sqrt{2}}{\sqrt{17}}\right] \\ & \therefore \int_0^2 \frac{2^x}{2^x\left(1+4^x\right)} \cdot d x=\frac{1}{(\log 2)} \cdot\left[\log \frac{4 \sqrt{2}}{\sqrt{17}}\right] \\ & =\log _2\left(\frac{4 \sqrt{2}}{\sqrt{17}}\right)\end{aligned}$

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