Definite Integrals — Maths STD 12 Science — Question

2. $\log2$

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 $\text{x}\rightarrow\infty;\text{ t}\rightarrow\infty$

$\therefore\text{I}=\int\limits^\infty_1\frac{1}{\text{t}(1+\text{t})}\text{dt}$

$=\int\limits^\infty_1\frac{1}{\text{t}+\text{t}^2}\text{dt}$

$=\int\limits^\infty_1\frac{1}{\big(\text{t}+\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}\text{dt}$

$=\frac{1}{2\times\frac{1}{2}}\Bigg[\log\Bigg|\frac{\text{t}+\frac{1}{2}-\frac{1}{2}}{\text{t}+\frac{1}{2}+\frac{1}{2}}\Bigg|\Bigg]^\infty_1$

$=\Big[\log\Big|\frac{\text{t}}{\text{t+1}}\Big|\Big]^\infty_1$

$=\Bigg[\log\Bigg|\frac{\frac{\text{t}}{\text{t}}}{\frac{\text{t}}{\text{t}}+\frac{1}{\text{t}}}\Bigg|\Bigg]^\infty_1$

$=\Bigg[\log\Bigg|\frac{1}{1+\frac{1}{\text{t}}}\Bigg|\Bigg]^\infty_1$

$=\log\frac{1}{1+0}-\log\frac{1}{1+1}$

$=\log(1)-\log(\frac{1}{2})$

$=0-(-\log2)$

$=\log2$