## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{4} e^{\sqrt{(4-x) /(4+x)}} \cdot \frac{d x}{(4+x) \sqrt{16-x^{2}}}
$$
## Solution

$$
\int_{0}^{4} e^{\sqrt{(4-x) /(4+x)}} \cdot \frac{d x}{(4+x) \sqrt{16-x^{2}}}=
$$

Let's make the substitution:

$$
\begin{aligned}
& t=\sqrt{\frac{4-x}{4+x}} \\
& d t=\frac{1}{2 \sqrt{\frac{4-x}{4+x}}} \cdot \frac{-4-x-4+x}{(4+x)^{2}} d x=-4 \sqrt{\frac{4+x}{4-x}} \cdot \frac{d x}{(4+x)^{2}}=-4 \frac{4+x}{\sqrt{16-x^{2}}} \cdot \frac{d x}{(4+x)^{2}}=\frac{-4 d x}{\sqrt{16-x^{2}} \cdot(4+x)} \\
& x=0 ; t=1 \\
& x=4 ; t=0
\end{aligned}
$$

We get:

$$
=\int_{1}^{0} e^{t} \cdot\left(-\frac{1}{4}\right) d t=\frac{1}{4} \int_{0}^{1} e^{t} d t=\left.\frac{1}{4} \cdot e^{t}\right|_{0} ^{1}=\frac{1}{4} \cdot(e-1)
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\�\�\�\�\� \%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5\%D0\%B3\%D1\%80\%D0\%B0\%D0\%BB $\% \mathrm{D} 1 \% 8 \mathrm{~B} \_11-14$ »

Categories: Kuznetsov Problem Book Integrals Problem 11 | Integrals

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## Problem Kuznetsov Integrals 11-15

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