Question

What is \(\rm \displaystyle\int_0^1 \frac{{{e^{{{\tan }^{ - 1}}x}}dx}}{{1 + {x^2}}}\) equal to ?
1. \({e^{\frac{\pi }{4}}} - 1\)
2. \({e^{\frac{\pi }{4}}} + 1\)
3.  \({e^{\frac{\pi }{2}}} + 1\)
4. e
5. e – 1

Answer 1

Correct Answer - Option 1 : \({e^{\frac{\pi }{4}}} - 1\)

Concept:

Integral Property:

\(\rm \displaystyle\int e^{x}\;dx = e^{x} + c\)

Calculation:

\(\rm I = \displaystyle \int\limits_{0}^{1}{\frac{e^{\tan^{-1}x}dx}{1+x^{2}}}\)

Let tan^-1 x = t

Differentiating both sides, we get

\(⇒\rm \frac{dx}{1+x^{2}} = dt\)

\(\rm I = \displaystyle\int\limits_{0}^{\frac{π}{4}} e^{t}dt\)

\(\rm ⇒ I = [{e^{t}}]_{0}^{\frac{π}{4}}\)

\(\rm \Rightarrow I = e^{\frac{\pi}{4}} - e^0\)

\(\rm \Rightarrow I = e^{\frac{\pi}{4}} - 1\)