## Problem Statement

Calculate the definite integral:

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

Let's introduce the substitution:

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

Then

$$
x=\frac{2-2 t^{2}}{2 t^{2}+1} \Rightarrow 2 x+2=\frac{6}{2 t^{2}+1} \Rightarrow d x=-\frac{12 t}{\left(2 t^{2}+1\right)} d t
$$

When $x=0, t=1$

When $x=2, t=0$

We get:

$$
\begin{aligned}
& \int_{0}^{2} \frac{(4 \sqrt{2-x}-\sqrt{2 x+2}) d x}{(\sqrt{2 x+2}+4 \sqrt{2-x})(2 x+2)^{2}}=-\int_{1}^{0} \frac{(4 t-1) \frac{12 t}{\left(2 t^{2}+1\right)^{2}}}{(1+4 t) \frac{36}{\left(2 t^{2}+1\right)^{2}}} d t= \\
& =-\int_{1}^{0} \frac{(4 t-1) \cdot 12 t}{(1+4 t) \cdot 36} d t=\frac{1}{3} \int_{0}^{1} \frac{(4 t-1) t}{4 t+1} d t=\frac{1}{3} \int_{0}^{1} \frac{4 t^{2}-t}{4 t+1} d t=
\end{aligned}
$$

The integrand is an improper fraction. Let's separate the integer part:

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

We get:

$$
\begin{aligned}
& =\frac{1}{3} \int_{0}^{1} \frac{4 t^{2}-t}{4 t+1} d t=\frac{1}{3} \int_{0}^{1}\left(\left(t-\frac{1}{2}\right)+\frac{\frac{1}{2}}{4 t+1}\right) d t=\frac{1}{3} \int_{0}^{1}\left(t-\frac{1}{2}+\frac{1}{8 t+2}\right)= \\
& =\frac{1}{3}\left(\int_{0}^{1} t d t-\frac{1}{2} \int_{0}^{1} d t+\int_{0}^{1} \frac{1}{8 t+2} d t\right)=\frac{1}{3}\left(\left.\frac{t^{2}}{2}\right|_{0} ^{1}-\left.\frac{1}{2} t\right|_{0} ^{1}+\frac{1}{2} \int_{0}^{1} \frac{1}{4 t+1} d t\right)= \\
& =\frac{1}{3}\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{8} \int_{0}^{1} \frac{1}{4 t+1} d(4 t+1)\right)=\frac{1}{3}\left(\left.\frac{1}{8} \ln |4 t+1|\right|_{0} ^{1}\right)=\frac{1}{24} \ln 5
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+11-12 »$ Categories: Kuznetsov's Problem Book Integrals Problem 11 | Integrals

- Last edited: 15:07, 26 June 2009.
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## Problem Kuznetsov Integrals 11-13

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