## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{7} \frac{\sqrt{x+25}}{(x+25)^{2} \sqrt{x+1}} d x
$$
## Solution

$$
\int_{0}^{7} \frac{\sqrt{x+25}}{(x+25)^{2} \sqrt{x+1}} d x=\int_{0}^{7} \frac{1}{(x+25) \sqrt{(x+25)(x+1)}} d x=
$$

Substitution:

$$
\begin{aligned}
& t=\sqrt{\frac{x+25}{x+1}} \\
& d t=\frac{1}{2} \sqrt{\frac{x+1}{x+25}} \cdot \frac{1 \cdot(x+1)-(x+25) \cdot 1}{(x+1)^{2}} d x=\frac{1}{2} \sqrt{\frac{x+1}{x+25}} \cdot \frac{-24}{(x+1)^{2}} d x= \\
& =-\frac{d x}{2(x+1) \sqrt{(x+25)(x+1)}} d x \Rightarrow \frac{d t}{(x+1) \sqrt{(x+25)(x+1)}}=-\frac{24}{12} \\
& x=0 \Rightarrow t=\sqrt{\frac{0+25}{0+1}}=\sqrt{25}=5 \\
& x=7 \Rightarrow t=\sqrt{\frac{7+25}{7+1}}=\sqrt{\frac{32}{8}}=\sqrt{4}=2
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int_{5}^{2} \frac{1}{\frac{x+25}{x+1} \cdot(x+1) \sqrt{(x+25)(x+1)}} d x=\int_{5}^{2} \frac{1}{t^{2}} \cdot\left(-\frac{d t}{12}\right)=\int_{5}^{2} \frac{-d t}{12 t^{2}}= \\
& =\left.\frac{1}{12 t}\right|_{5} ^{2}=\frac{1}{12 \cdot 2}-\frac{1}{12 \cdot 5}=\frac{5-2}{12 \cdot 2 \cdot 5}=\frac{1}{40}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+11-18$ » Categories: Kuznetsov's Problem Book Integrals Problem 11 | Integrals

- Last edited: 07:18, 26 June 2009.
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## Problem Kuznetsov Integrals 11-19

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