## Problem Statement

Calculate the definite integral:

$$
\int_{1 / 24}^{1 / 3} \frac{5 \sqrt{x+1}}{(x+1)^{2} \sqrt{x}} d x
$$
## Solution

$$
\int_{1 / 24}^{1 / 3} \frac{5 \sqrt{x+1}}{(x+1)^{2} \sqrt{x}} d x=\int_{1 / 24}^{1 / 3} \frac{5}{(x+1) \sqrt{(x+1) x}} d x=
$$

Substitution:

$$
\begin{aligned}
& t=\sqrt{\frac{x+1}{x}} \\
& d t=\frac{1}{2} \sqrt{\frac{x}{x+1}} \cdot \frac{1 \cdot x-(x+1) \cdot 1}{x^{2}} d x=\frac{1}{2} \sqrt{\frac{x}{x+1}} \cdot \frac{-1}{x^{2}} d x= \\
& =-\frac{1}{2 x \sqrt{(x+1) x}} d x \Rightarrow \frac{d x}{x \sqrt{(x+1) x}}=-2 d t \\
& x=\frac{1}{24} \Rightarrow t=\sqrt{\frac{\frac{1}{24}+1}{\frac{1}{24}}}=\sqrt{1+24}=5 \\
& x=\frac{1}{3} \Rightarrow t=\sqrt{\frac{\frac{1}{3}+1}{\frac{1}{3}}}=\sqrt{1+3}=2
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int_{1 / 24}^{1 / 3} \frac{5}{\frac{x+1}{x} \cdot x \sqrt{(x+1) x}} d x=\int_{5}^{2} \frac{5}{t^{2}} \cdot(-2 d t)=\int_{5}^{2} \frac{-10 d t}{t^{2}}= \\
& =\left.\frac{10}{t}\right|_{5} ^{2}=\frac{10}{2}-\frac{10}{5}=5-2=3
\end{aligned}
$$

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}+11-22 »

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

- Last modified: 15:27, 26 June 2009.
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## Problem Kuznetsov Integrals 11-23

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