## Problem Statement

Find the indefinite integral:

$$
\int \frac{\sqrt{1+x}}{x^{2} \cdot \sqrt{x}} d x
$$
## Solution

$$
\int \frac{\sqrt{1+x}}{x^{2} \cdot \sqrt{x}} d x=\int x^{-\frac{5}{2}}(1+x)^{\frac{1}{2}} d x=
$$

Under the integral is a binomial differential $x^{m}\left(a+b x^{n}\right)^{p}$, from which

$$
m=-\frac{5}{2} ; n=1 ; p=\frac{1}{2}
$$

Since $\frac{m+1}{n}+p=\frac{-\frac{5}{2}+1}{1}+\frac{1}{2}=-\frac{3}{2}+\frac{1}{2}=-1$ is an integer, we use the substitution:

$$
a x^{-n}+b=z^{s}, \text { where } s \text { is the denominator of the fraction } p .
$$

In our case, the substitution is:

$$
\begin{aligned}
& x^{-1}+1=z^{2} \\
& x=\left(z^{2}-1\right)^{-1} \\
& d x=-1 \cdot\left(z^{2}-1\right)^{-2} \cdot 2 z d z=-2 z\left(z^{2}-1\right)^{-2} d z
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int\left(\left(z^{2}-1\right)^{-1}\right)^{-\frac{5}{2}}\left(1+\left(z^{2}-1\right)^{-1}\right)^{\frac{1}{2}}(-2 z)\left(z^{2}-1\right)^{-2} d z= \\
& =-2 \cdot \int\left(z^{2}-1\right)^{\frac{5}{2}}\left(1+\frac{1}{z^{2}-1}\right)^{\frac{1}{2}} z\left(z^{2}-1\right)^{-2} d z= \\
& =-2 \cdot \int\left(z^{2}-1\right)^{\frac{1}{2}}\left(\frac{z^{2}}{z^{2}-1}\right)^{\frac{1}{2}} z d z=-2 \cdot \int\left(z^{2}-1\right)^{\frac{1}{2}} \frac{z}{\left(z^{2}-1\right)^{\frac{1}{2}}} \cdot z d z= \\
& =-2 \cdot \int z^{2} d z=-\frac{2 z^{3}}{3}+C=-\frac{2\left(\sqrt{x^{-1}+1}\right)^{3}}{3}+C=
\end{aligned}
$$

$$
=-\frac{2\left(\sqrt{\frac{1}{x}+1}\right)^{3}}{3}+C=-\frac{2}{3}\left(\sqrt{\frac{1+x}{x}}\right)^{3}+C
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+13-10$ » Categories: Kuznetsov's Problem Book Integrals Problem $13 \mid$ Integrals

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## Problem Kuznetsov Integrals 13-11

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