## Problem Statement

Calculate the definite integral:

$$
\int_{1 / 8}^{1} \frac{15 \sqrt{x+3}}{(x+3)^{2} \sqrt{x}} d x
$$
## Solution

$$
\int_{1 / 8}^{1} \frac{15 \sqrt{x+3}}{(x+3)^{2} \sqrt{x}} d x=\int_{1 / 8}^{1} \frac{15}{(x+3) \sqrt{(x+3) x}} d x=
$$

Substitution:

$$
\begin{aligned}
& t=\sqrt{\frac{x+3}{x}} \\
& d t=\frac{1}{2} \sqrt{\frac{x}{x+3}} \cdot \frac{1 \cdot x-(x+3) \cdot 1}{x^{2}} d x=\frac{1}{2} \sqrt{\frac{x}{x+3}} \cdot \frac{-3}{x^{2}} d x= \\
& =-\frac{3}{2 x \sqrt{(x+3) x}} d x \Rightarrow \frac{d x}{x \sqrt{(x+3) x}}=-\frac{2 d t}{3} \\
& x=\frac{1}{8} \Rightarrow t=\sqrt{\frac{\frac{1}{8}+3}{\frac{1}{8}}}=\sqrt{1+24}=5 \\
& x=1 \Rightarrow t=\sqrt{\frac{1+3}{1}}=\sqrt{4}=2
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int_{1 / 8}^{1} \frac{15}{\frac{x+3}{x} \cdot x \sqrt{(x+3) x}} d x=\int_{5}^{2} \frac{15}{t^{2}} \cdot\left(-\frac{2 d t}{3}\right)=\int_{5}^{2} \frac{-10 d t}{t^{2}}= \\
& =\left.\frac{10}{t}\right|_{5} ^{2}=\frac{10}{2}-\frac{10}{5}=5-2=3
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_ $\% \mathrm{D} 0 \% 9 \mathrm{~A} \% \mathrm{D} 1 \% 83 \% \mathrm{D} 0 \% \mathrm{~B} 7 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 1 \% 86 \% \mathrm{D} 0 \% \mathrm{BE} \% \mathrm{D} 0 \% \mathrm{~B} 2 \mathrm{\%} 0 \mathrm{D} 0 \% 98 \% \mathrm{D} 0 \% \mathrm{BD}$ $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+11-15$ »

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

- Last modified: 21:56, 18 June 2009.
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## Problem Kuznetsov Integrals 11-16

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