## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{1} \frac{(4 \sqrt{1-x}-\sqrt{x+1}) d x}{(\sqrt{x+1}+4 \sqrt{1-x})(x+1)^{2}}
$$
## Solution

$$
\int_{0}^{1} \frac{(4 \sqrt{1-x}-\sqrt{x+1}) d x}{(\sqrt{x+1}+4 \sqrt{1-x})(x+1)^{2}}=\int_{0}^{1} \frac{\left(4 \sqrt{\frac{1-x}{x+1}}-1\right) d x}{\left(1+4 \sqrt{\frac{1-x}{x+1}}\right)(x+1)^{2}}=
$$

Introduce the substitution:

$$
t=\sqrt{\frac{1-x}{x+1}}
$$

Then

$$
\begin{aligned}
& t^{2}=\frac{1-x}{x+1}=-\frac{x+1-2}{x+1}=-1+\frac{2}{x+1} \Rightarrow \frac{2}{t^{2}+1}=x+1 \Rightarrow \\
& x=\frac{1-t^{2}}{t^{2}+1} \Rightarrow d x=\frac{(-2 t) \cdot\left(t^{2}+1\right)-\left(1-t^{2}\right) \cdot 2 t}{\left(t^{2}+1\right)^{2}} d t \Rightarrow \\
& d x=\frac{-4 t}{\left(t^{2}+1\right)^{2}} d t \\
& \text { When } x=0, t=\sqrt{\frac{1-0}{0+1}}=1 \\
& \text { When } x=1, t=\sqrt{\frac{1-1}{1+1}}=0
\end{aligned}
$$

We get:

$$
=\int_{1}^{0} \frac{4 t-1}{(1+4 t) \cdot \frac{4}{\left(t^{2}+1\right)^{2}}} \cdot \frac{-4 t}{\left(t^{2}+1\right)^{2}} d t=\int_{0}^{1} \frac{(4 t-1) t}{1+4 t} d t=\int_{0}^{1} \frac{4 t^{2}-t}{4 t+1} d t=
$$

The integrand is an improper fraction. We separate the integer part:

$$
\begin{array}{ll}
4 t^{2}-t & \mid 4 t+1 \\
\frac{4 t^{2}+t}{-2 t} & t-\frac{1}{2} \\
\frac{-2 t-\frac{1}{2}}{\frac{1}{2}}
\end{array}
$$

We get:

$$
\begin{aligned}
& =\int_{0}^{1}\left(\left(t-\frac{1}{2}\right)+\frac{\frac{1}{2}}{4 t+1}\right) d t=\int_{0}^{1}\left(t-\frac{1}{2}+\frac{1}{8 t+2}\right)= \\
& =\int_{0}^{1} t d t-\frac{1}{2} \int_{0}^{1} d t+\int_{0}^{1} \frac{1}{8 t+2} d t=\left.\frac{t^{2}}{2}\right|_{0} ^{1}-\left.\frac{1}{2} t\right|_{0} ^{1}+\frac{1}{2} \int_{0}^{1} \frac{1}{4 t+1} d t= \\
& =\frac{1}{2}-\frac{1}{2}+\frac{1}{8} \int_{0}^{1} \frac{1}{4 t+1} d(4 t+1)=\left.\frac{1}{8} \ln |4 t+1|\right|_{0} ^{1}=\frac{1}{8} \ln 5
\end{aligned}
$$

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%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_%D0%97%D0%B0%D0%B4%D0%B0%D1%87%D0%B0_11-29" Categories: Kuznetsov's Integral Problems Problem 11 | Integrals

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