## Problem Statement

Calculate the definite integral:

$$
\int_{5 / 2}^{10 / 3} \frac{\sqrt{x+2}+\sqrt{x-2}}{(\sqrt{x+2}-\sqrt{x-2})(x-2)^{2}} d x
$$
## Solution

$$
\int_{5 / 2}^{10 / 3} \frac{\sqrt{x+2}+\sqrt{x-2}}{(\sqrt{x+2}-\sqrt{x-2})(x-2)^{2}} d x=
$$

Divide the numerator and the denominator by $\sqrt{x-2}$:

$$
=\int_{5 / 2}^{10 / 3} \frac{\sqrt{\frac{x+2}{x-2}}+1}{\left(\sqrt{\frac{x+2}{x-2}}-1\right)(x-2)^{2}} d x=
$$

Substitution:

$$
\begin{aligned}
& t=\sqrt{\frac{x+2}{x-2}} \Rightarrow t^{2}=\frac{x+2}{x-2}=1+\frac{4}{x-2} \Rightarrow \frac{4}{t^{2}-1}=x-2 \Rightarrow x=\frac{2 t^{2}+2}{t^{2}-1} \\
& d x=\frac{4 t \cdot\left(t^{2}-1\right)-\left(2 t^{2}+2\right) \cdot 2 t}{\left(t^{2}-1\right)^{2}} d t=\frac{4 t^{3}-4 t-4 t^{3}-4 t}{\left(t^{2}-1\right)^{2}} d t=\frac{-8 t}{\left(t^{2}-1\right)^{2}} d t \\
& x=\frac{5}{2} \Rightarrow t=\sqrt{\frac{\frac{5}{2}+2}{\frac{5}{2}-2}}=\sqrt{\frac{5+4}{5-4}}=\sqrt{9}=3 \\
& x=\frac{10}{3} \Rightarrow t=\sqrt{\frac{\frac{10}{3}+2}{\frac{10}{3}-2}}=\sqrt{\frac{10+6}{10-6}}=\sqrt{4}=2
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int_{3}^{2} \frac{t+1}{(t-1)\left(\frac{4}{t^{2}-1}\right)^{2}} \cdot \frac{-8 t}{\left(t^{2}-1\right)^{2}} d t=\int_{3}^{2} \frac{t+1}{(t-1) \cdot \frac{16}{\left(t^{2}-1\right)^{2}}} \cdot \frac{-8 t}{\left(t^{2}-1\right)^{2}} d t= \\
& =\frac{1}{2} \cdot \int_{3}^{2} \frac{(t+1) t}{t-1} d t=\frac{1}{2} \cdot \int_{3}^{2} \frac{(t-1) t+(2 t-2)+2}{t-1} d t=
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{1}{2} \cdot \int_{3}^{2}\left(t+2+\frac{2}{t-1}\right) d t=\left.\frac{1}{2} \cdot\left(\frac{t^{2}}{2}+2 t+2 \ln |t-1|\right)\right|_{3} ^{2}= \\
& =\frac{1}{2} \cdot\left(\frac{3^{2}}{2}+2 \cdot 3+2 \ln |3-1|\right)-\frac{1}{2} \cdot\left(\frac{2^{2}}{2}+2 \cdot 2+2 \ln |2-1|\right)= \\
& =\frac{1}{2} \cdot\left(\frac{9}{2}+6+2 \ln 2\right)-\frac{1}{2} \cdot(2+4+2 \ln 1)=\frac{9}{4}+3+\ln 2-3=\frac{9}{4}+\ln 2
\end{aligned}
$$

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

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

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