Example 6. Find the integral $\int \frac{d x}{\sqrt[4]{1+x^{4}}}$.
Solution. Rewriting the integral as

$$
\int \frac{d x}{\sqrt[4]{1+x^{4}}}=\int x^{0}\left(1+x^{4}\right)^{-\frac{1}{4}} d x
$$

we conclude that $m=0, n=4, p=-\frac{1}{4}$. Since $\frac{m+1}{n}+p=\frac{1}{4}-\frac{1}{4}=0$ (an integer), we have the third case of integrability. The substitution

$$
a x^{-n}+b=t^{s}
$$

where $s$ is the denominator of the number $p$, in this case will be

$$
x^{-4}+1=t^{4}
$$

from which

$$
\begin{gathered}
x=\left(t^{4}-1\right)^{-\frac{1}{4}}, d x=-t^{3}\left(t^{4}-1\right)^{-\frac{5}{4}} d t \\
\sqrt[4]{1+x^{4}}=t x=t\left(t^{4}-1\right)^{-\frac{1}{4}}
\end{gathered}
$$

Therefore,

$$
\int \frac{d x}{\sqrt[4]{1+x^{4}}}=-\int \frac{t^{2} d t}{t^{4}-1}=\frac{1}{4} \int\left(\frac{1}{t+1}-\frac{1}{t-1}\right) d t-
$$

$$
-\frac{1}{2} \int \frac{d t}{t^{2}+1}=\frac{1}{4} \ln \left|\frac{t+1}{t-1}\right|-\frac{1}{2} \operatorname{arctg} t+C
$$

## Problems

Find the integrals:

1. $\int \frac{\sqrt[6]{2 x-1}+1}{(2 x-1)(\sqrt[3]{2 x-1}-1)} d x$.
2. $\int \frac{d x}{\sqrt[4]{(x-1)^{3}(x+2)^{5}}}$.
3. $\int \frac{d x}{\sqrt[3]{(x-1)(x+1)^{2}}}$.
4. $\int \frac{x+\sqrt[3]{x^{2}}+\sqrt[6]{x}}{x(1-\sqrt[3]{x})} d x$.
5. $\int \frac{d x}{x \sqrt[3]{1+x^{5}}}$.
6. $\int \frac{d x}{\sqrt[3]{x^{2}}(1+\sqrt{x})^{2}}$.
7. $\int x^{-6}\left(1+x^{2}\right)^{\frac{1}{2}} d x$.
8. $\int \frac{d x}{\sqrt[3]{x^{2}}\left(1+\sqrt[3]{x^{2}}\right)}$.

## Answers

1. $3 \ln |\sqrt[6]{2 x-1}-1|-3 \ln \sqrt[6]{2 x-1}+C$.
2. $\frac{4}{3} \sqrt[4]{\frac{x-1}{x+2}}+C . \quad$ Hint. Transform the integrand function $\sqrt[4]{(x-1)^{3}(x+2)^{5}}=$

$$
\begin{aligned}
& =(x-1)(x+2) \sqrt[4]{\frac{x+2}{x-1}} \\
& \text { 3. } \frac{1}{2} \ln \frac{t^{2}+t+1}{(t-1)^{2}}+\sqrt{3} \operatorname{arctg} \frac{2 t+1}{\sqrt{3}}+C \\
& t=\sqrt[3]{\frac{x+1}{x-1}} \\
& \text { 4. } \frac{3}{2} \sqrt[3]{x^{2}}+6 \operatorname{arctg} \sqrt[6]{x}+C \\
& \text { 5. } \frac{1}{10} \ln \frac{(t-1)^{2}}{t^{2}+t+1}+ \\
& +\frac{\sqrt{3}}{5} \operatorname{arctg} \frac{2 t+1}{\sqrt{3}}+C, t=\sqrt[3]{1+x^{5}} \\
& \text { 6. }-\frac{3}{\sqrt[3]{x}+1}+C \\
& \text { 7. }-\frac{1}{5}\left(\frac{1+x^{2}}{x^{2}}\right)^{\frac{5}{2}}+ \\
& +\frac{1}{3}\left(\frac{1+x^{2}}{x^{2}}\right)^{\frac{3}{2}}+C .8 .3 \operatorname{arctg} \sqrt[3]{x}+C .
\end{aligned}
$$

## § 4.9. Integration of Hyperbolic Functions

The integration of hyperbolic functions is based on the formulas:

$$
\int \operatorname{ch} x d x=\operatorname{sh} x+C
$$

$$
\begin{aligned}
& \int \operatorname{sh} x d x=\operatorname{ch} x+C \\
& \int \frac{d x}{\operatorname{ch}^{2} x}=\operatorname{th} x+C \\
& \int \frac{d x}{\operatorname{sh}^{2} x}=-\operatorname{cth} x+C
\end{aligned}
$$

Integrals of even powers of $\operatorname{sh} x$ and $\operatorname{ch} x$ are found using the formulas:

$$
\operatorname{ch}^{2} x=\frac{1}{2}(\operatorname{ch} 2 x+1) ; \operatorname{sh}^{2} x=\frac{1}{2}(\operatorname{ch} 2 x-1) ; \operatorname{sh} x \operatorname{ch} x=\frac{\operatorname{sh} 2 x}{2} \text {. }
$$

Integrals of odd powers of $\operatorname{sh} x$ and $\operatorname{ch} x$ are found by separating a first-degree factor and introducing a new variable.