## Solution

Introduce the substitution:

$$
t=\sqrt{\frac{2-x}{3 x+2}}
$$

Then:

$$
\begin{aligned}
& x=\frac{2-2 t^{2}}{3 t^{2}+1}, \quad d x=-\frac{16 t}{\left(3 t^{2}+1\right)^{2}} d t \\
& \text { When } x=0, \quad t=1 \\
& \text { When } x=2, \quad t=0 \\
& 3 x+2=\frac{8}{3 t^{2}+1}
\end{aligned}
$$

We obtain:

$$
\int_{0}^{2} \frac{(4 \sqrt{2-x}-\sqrt{3 x+2}) d x}{(\sqrt{3 x+2}+4 \sqrt{2-x})(3 x+2)^{2}}=-\int_{1}^{0} \frac{(4 t-1) \frac{16 t}{\left(3 t^{2}+1\right)^{2}}}{(1+4 t) \frac{64}{\left(3 t^{2}+1\right)^{2}}} d t=\frac{1}{4} \int_{0}^{1} \frac{4 t^{2}-t}{1+4 t} d t=
$$

The integrand is an improper fraction. 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:

$$
=\frac{1}{4} \int_{0}^{1} t d t-\frac{1}{8} \int_{0}^{1} d t+\frac{1}{8} \int_{0}^{1} \frac{d t}{1+4 t}=\frac{1}{8}-\frac{1}{8}+\left.\frac{1}{32} \ln |1+4 t|\right|_{0} ^{1}=\frac{1}{32} \ln 5
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+11-19$ »

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

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## Problem Kuznetsov Integrals 11-20

## Material from PlusPi
## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{2} \frac{(4 \sqrt{2-x}-\sqrt{3 x+2}) d x}{(\sqrt{3 x+2}+4 \sqrt{2-x})(3 x+2)^{2}}
$$