## Solution

$$
\int_{0}^{2} e^{\sqrt{(2-x) /(2+x)}} \cdot \frac{d x}{(2+x) \sqrt{4-x^{2}}}=
$$

Substitution:

$$
\begin{aligned}
& t=\sqrt{\frac{2-x}{2+x}} \\
& d t=\frac{1}{2} \cdot \sqrt{\frac{2+x}{2-x}} \cdot\left(\frac{2-x}{2+x}\right)^{\prime} \cdot d x=\frac{1}{2} \cdot \sqrt{\frac{2+x}{2-x}} \cdot \frac{-4}{(2+x)^{2}} \cdot d x=-\frac{2 d x}{\sqrt{4-x^{2}}(2+x)} \\
& x=0 \Rightarrow t=\sqrt{\frac{2-0}{2+0}}=1 \\
& x=2 \Rightarrow t=\sqrt{\frac{2-2}{2+2}}=0
\end{aligned}
$$

We get:

$$
=\int_{1}^{0} e^{t} \cdot\left(\frac{-d t}{2}\right)=-\frac{1}{2} \cdot \int_{1}^{0} e^{t} d t=-\left.\frac{e^{t}}{2}\right|_{1} ^{0}=-\frac{e^{0}}{2}+\frac{e^{1}}{2}=\frac{e-1}{2}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+11-20$ »

Categories: Kuznetsov's Problem Book Integrals Problem 11 | Integrals

- Last edited: 15:18, 26 June 2009.
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## Problem Kuznetsov Integrals 11-21

## Material from PlusPi
## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{2} e^{\sqrt{(2-x) /(2+x)}} \cdot \frac{d x}{(2+x) \sqrt{4-x^{2}}}
$$