Integrals — Class 12 Maths Solution

Let $I = \int_0^1 {\frac{{dx}}{{{e^x} + {e^{ - x}}}}} = \int_0^1 {\frac{{{e^x}}}{{1 + {e^{2x}}}}} dx$

Let's put ${e^x} = t$

$\Rightarrow$ ${e^x}dx = dt$

therefore,$I = \int_1^e {\frac{{dt}}{{1 + {t^2}}}} = \left[ {{{\tan }^{ - 1}}t} \right]_1^e$

$= {\tan ^{ - 1}}e - {\tan ^{ - 1}}1$

$= {\tan ^{ - 1}}e - \frac{\pi }{4}$