## Solution

$$
\int_{0}^{6} \frac{e^{\sqrt{(6-x) /(6+x)}} \cdot d x}{(6+x) \sqrt{36-x^{2}}}=\int_{0}^{6} \frac{e^{\sqrt{(6-x) /(6+x)}} \cdot d x}{(6+x)^{2} \sqrt{\frac{6-x}{6+x}}}=-\left.\frac{1}{6} \cdot e^{\sqrt{\frac{6-x}{6+x}}}\right|_{0} ^{6}=\frac{e-1}{6}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+11-27$ "

Categories: Kuznetsov Problem Book Integrals Problem 11 | Integrals

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## Problem Kuznetsov Integrals 11-28

## Material from PlusPi
## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{6} \frac{e^{\sqrt{(6-x) /(6+x)}} \cdot d x}{(6+x) \sqrt{36-x^{2}}}
$$