^ _01 1+e^x dx equals: 1. 2-1 2. 2 3. 4-1 4. - 2 - Vidyadip

Question

$\int\limits^\infty_0\frac{1}{1+\text{e}^\text{x}}\text{dx}$ equals:

$\log2-1$
$\log2$
$\log4-1$
$-\log2$

✓

Answer

$\log2$

Solution:
We have,
$\text{I}=\int\limits^\infty_0\frac{1}{1+\text{e}^\text{x}}\text{ dx}$
Putting $\text{e}^\text{x}=\text{t}$
$\Rightarrow \text{e}^\text{x}\text{ dx}=\text{dt}$
$\Rightarrow \text{dx} = \frac{\text{dt}}{\text{t}}$
When $\text{x}\rightarrow0;\text{ t}\rightarrow1$
and $\text{x}\rightarrow\infty;\text{ t}\rightarrow\infty$
$\therefore\text{I}=\int\limits^\infty_1\frac{1}{\text{t}(1+\text{t})}\text{dt}$
$=\int\limits^\infty_1\frac{1}{\text{t}+\text{t}^2}\text{dt}$
$=\int\limits^\infty_1\frac{1}{\big(\text{t}+\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}\text{dt}$
$=\frac{1}{2\times\frac{1}{2}}\Bigg[\log\Bigg|\frac{\text{t}+\frac{1}{2}-\frac{1}{2}}{\text{t}+\frac{1}{2}+\frac{1}{2}}\Bigg|\Bigg]^\infty_1$
$=\Big[\log\Big|\frac{\text{t}}{\text{t+1}}\Big|\Big]^\infty_1$
$=\Bigg[\log\Bigg|\frac{\frac{\text{t}}{\text{t}}}{\frac{\text{t}}{\text{t}}+\frac{1}{\text{t}}}\Bigg|\Bigg]^\infty_1$
$=\Bigg[\log\Bigg|\frac{1}{1+\frac{1}{\text{t}}}\Bigg|\Bigg]^\infty_1$
$=\log\frac{1}{1+0}-\log\frac{1}{1+1}$
$=\log(1)-\log(\frac{1}{2})$
$=0-(-\log2)$
$=\log2$

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