## Problem Statement

Calculate the definite integral:

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

$$
\int_{0}^{2} \frac{(4 \sqrt{2-x}-\sqrt{x+2}) d x}{(\sqrt{x+2}+4 \sqrt{2-x})(x+2)^{2}}=\int_{0}^{2} \frac{\left(4 \sqrt{\frac{2-x}{x+2}}-1\right) d x}{\left(1+4 \sqrt{\frac{2-x}{x+2}}\right)(x+2)^{2}}=
$$

Introduce the substitution:

$$
t=\sqrt{\frac{2-x}{x+2}}
$$

Then

$$
\begin{aligned}
& t^{2}=\frac{2-x}{x+2}=-\frac{x+2-4}{x+2}=-1+\frac{4}{x+2} \Rightarrow \frac{4}{t^{2}+1}=x+2 \Rightarrow \\
& x=\frac{2-2 t^{2}}{t^{2}+1} \Rightarrow d x=\frac{(-4 t) \cdot\left(t^{2}+1\right)-\left(2-2 t^{2}\right) \cdot 2 t}{\left(t^{2}+1\right)^{2}} d t \Rightarrow \\
& d x=\frac{-8 t}{\left(t^{2}+1\right)^{2}} d t \\
& \text { When } x=0, t=\sqrt{\frac{2-0}{0+2}}=1 \\
& \text { When } x=2, t=\sqrt{\frac{2-2}{2+2}}=0
\end{aligned}
$$

We get:

$$
=\int_{1}^{0} \frac{4 t-1}{(1+4 t) \cdot \frac{4}{\left(t^{2}+1\right)^{2}}} \cdot \frac{-8 t}{\left(t^{2}+1\right)^{2}} d t=2 \cdot \int_{0}^{1} \frac{(4 t-1) t}{1+4 t} d t=2 \cdot \int_{0}^{1} \frac{4 t^{2}-t}{4 t+1} d t=
$$

The integrand is an improper fraction. We separate the integer part:

$$
\begin{array}{ll}
4 t^{2}-t & \mid 4 t+1 \\
\frac{4 t^{2}+t}{-2 t} & t-\frac{1}{2} \\
\frac{-2 t-\frac{1}{2}}{\frac{1}{2}}
\end{array}
$$

We get:

$$
\begin{aligned}
& =2 \cdot \int_{0}^{1}\left(\left(t-\frac{1}{2}\right)+\frac{\frac{1}{2}}{4 t+1}\right) d t=2 \cdot \int_{0}^{1}\left(t-\frac{1}{2}+\frac{1}{8 t+2}\right)= \\
& =2 \cdot \int_{0}^{1} t d t-\int_{0}^{1} d t+\int_{0}^{1} \frac{1}{4 t+1} d t=\left.2 \cdot \frac{t^{2}}{2}\right|_{0} ^{1}-\left.t\right|_{0} ^{1}+\int_{0}^{1} \frac{1}{4 t+1} d t= \\
& =1-1+\frac{1}{4} \int_{0}^{1} \frac{1}{4 t+1} d(4 t+1)=\left.\frac{1}{4} \ln |4 t+1|\right|_{0} ^{1}=\frac{1}{4} \ln 5
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+11-31 »$ Categories: Kuznetsov's Problem Book Integrals Problem 11 | Integrals

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- Last edited: 14:13, July 3, 2009.
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