## Problem Statement

Calculate the definite integral:

$$
\int_{16 / 15}^{4 / 3} \frac{4 \sqrt{x}}{x^{2} \sqrt{x-1}} d x
$$
## Solution

## Problem 11.

## Calculate the definite integral:

$$
I=\int_{16 / 15}^{4 / 3} \frac{4 \sqrt{x}}{x^{2} \sqrt{x-1}} d x
$$

## Perform a variable substitution:

$$
\begin{aligned}
& t=\sqrt{\frac{x}{x-1}}, \text { hence } t^{2}=\frac{x}{x-1}, \text { that is } x=1+\frac{1}{t^{2}-1}, \text { and } \\
& d x=\frac{-2 t}{\left(t^{2}-1\right)^{2}} d t=-\frac{2 t d t}{\left(t^{2}-1\right)^{2}} . \\
& x=\frac{1}{t^{2}-1}+1=\frac{t^{2}}{t^{2}-1}
\end{aligned}
$$

Recalculate the limits of integration

$$
\begin{aligned}
& t_{1}=\sqrt{\frac{4 / 3}{4 / 3-1}}=2 \\
& t_{2}=\sqrt{\frac{16 / 15}{16 / 15-1}}=4
\end{aligned}
$$

## Take the resulting integral:

$$
I=\int_{4}^{2} \frac{4 t}{\frac{t^{4}}{\left(t^{2}-1\right)^{2}}} \cdot\left(-\frac{2 t}{\left(t^{2}-1\right)^{2}} d t\right)=8 \int_{2}^{4} \frac{1}{t^{2}} d t=-\left.\frac{8}{t}\right|_{2} ^{4}=-2+4=2
$$

Answer: $\int_{16 / 15}^{4 / 3} \frac{4 \sqrt{x}}{x^{2} \sqrt{x-1}} d x=2$

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## Problem Kuznetsov Integrals 11-27

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