Example. Find the indefinite integral

$$
\int \frac{\sqrt[5]{1+\sqrt[3]{x}}}{x \sqrt[5]{x^{2}}} d x
$$
Solution. Rewrite the integral as

$$
\int \frac{\sqrt[5]{1+\sqrt[3]{x}}}{x \sqrt[5]{x^{2}}} d x=\int x^{-7 / 5}\left(1+x^{1 / 3}\right)^{1 / 5} d x
$$

The integrand has the form $x^{m}\left(1+x^{n}\right)^{p}$ for

$$
m=-\frac{7}{5}, \quad n=\frac{1}{3}, \quad p=\frac{1}{5}, \quad \frac{m+1}{n}+p=-1
$$

Therefore, the third case of integrability applies.

## Applying the substitution

$$
x^{-1 / 3}+1=z^{5}
$$

and noting that

$$
x=\left(z^{5}-1\right)^{-3}, \quad d x=-3\left(z^{5}-1\right)^{-4} 5 z^{4} d z
$$

we get

$$
\begin{aligned}
& \int x^{-7 / 5}\left(1+x^{1 / 3}\right)^{1 / 5} d x= \\
& =\int\left(z^{5}-1\right)^{21 / 5}\left(1+\left(z^{5}-1\right)^{-1}\right)^{1 / 5} \frac{(-15) z^{4}}{\left(z^{5}-1\right)^{4}} d z= \\
& =-\frac{15}{6} z^{6}+C=-\frac{5}{2}\left(x^{-1 / 3}+1\right)^{6 / 5}+C=-\frac{5}{2}\left(\frac{1+\sqrt[3]{x}}{\sqrt[3]{x}}\right)^{6 / 5}+C \\
& \text { Answer. } \int \frac{\sqrt[5]{1+\sqrt[3]{x}}}{x \sqrt[5]{x^{2}}} d x=-\frac{5}{2}\left(\frac{1+\sqrt[3]{x}}{\sqrt[3]{x}}\right)^{6 / 5}+C
\end{aligned}
$$

Conditions of the Problems. Find the indefinite integrals.

1. $\int \frac{\sqrt[3]{1+\sqrt[4]{x^{3}}}}{x^{2}} d x$.
2. $\quad \int \frac{\sqrt{1+x}}{x^{2} \sqrt{x}} d x$.
3. $\int \frac{1}{x \sqrt{1+x^{3}}} d x$.
4. $\int \frac{\sqrt{1-x^{4}}}{x^{5}} d x$.
5. $\int \frac{1}{x^{3} \sqrt[3]{2-x^{2}}} d x$.
6. $\int \frac{\sqrt[3]{1+\sqrt[4]{x}}}{\sqrt{x}} d x$.
7. $\int \frac{1}{x^{4} \sqrt{1+x^{2}}} d x$.
8. $\int \frac{1}{\sqrt{x}(\sqrt[4]{x}+1)^{10}} d x$.
9. $\int \frac{1}{x \sqrt[3]{(x+1)^{2}}} d x$.
10. $\int \frac{1}{\sqrt[4]{1+x^{4}}} d x$.

Answers.

1. $-\left(\frac{1+\sqrt[4]{x^{3}}}{\sqrt[4]{x^{3}}}\right)^{4 / 3}+C$
2. $-\frac{2}{3}\left(\frac{1+x}{x}\right)^{3 / 2}+C$.
3. $\frac{2}{3} \ln \left(\sqrt{1+x^{3}}-1\right)-\ln |x|+C$.
4. $\frac{1}{4} \ln \frac{1+\sqrt{1-x^{4}}}{x^{2}}-\frac{1}{4} \frac{\sqrt{1-x^{4}}}{x^{4}}+C$.
5. $-\frac{1}{4}\left(\frac{\sqrt[3]{2-x^{3}}}{x}\right)^{2}+$ C. $\quad$ 6. $\frac{3}{7}(4 \sqrt{x}+\sqrt[4]{x}-3) \sqrt[3]{1+\sqrt[4]{x}}+C$.
6. $\frac{\left(2 x^{2}-1\right) \sqrt{1+x^{2}}}{3 x^{3}}+C . \quad 8 . \quad-\frac{1}{2(\sqrt[4]{x}+1)^{8}}+\frac{4}{9(\sqrt[4]{x}+1)^{9}}+C$.
7. $\frac{3}{\sqrt[3]{x+1}}+\ln \frac{|x|}{(\sqrt[3]{x}+1)^{3}}+C$.
8. $\frac{1}{4} \ln \frac{\sqrt[4]{1+x^{4}}+x}{\sqrt[4]{1+x^{4}}-x}-\frac{1}{2} \operatorname{arctg} \frac{\sqrt[4]{1+x^{4}}}{x}+C$

## Chapter 8

## DEFINITE INTEGRAL

In the topic of DEFINITE INTEGRAL, you will learn about the Newton-Leibniz formula and how to apply it to compute definite integrals using techniques for finding antiderivatives. You will learn to apply definite integrals to solve geometric problems (calculating areas of plane figures, lengths of curve arcs, and volumes of solids).

Using the RESHEBNIK.VM package, you can find new limits of integration (when using various substitutions), compute antiderivatives, and apply the Newton-Leibniz formula, perform all numerical calculations, and verify the correctness of the results you obtained.

## 8.1. Integration by Substitution under the Differential Sign

Problem Statement. Compute the definite integral

$$
\int_{a}^{b} F(x) g(x) d x
$$

Plan of Solution. Suppose $g(x)$ has an obvious antiderivative $G(x)$, and $F(x)$ is a function of this antiderivative, i.e., $F(x)=u(G(x))$. Then

$$
\int_{a}^{b} F(x) g(x) d x=\int_{a}^{b} u(G(x)) G^{\prime}(x) d x=\int_{\alpha}^{\beta} u(G) d G
$$

where $\alpha=G(a), \beta=G(b)$.

This kind of transformation is called substitution under the differential sign.

If the method is chosen correctly, the last integral turns out to be a table integral or can be reduced to a table integral in a known way, after which we apply the Newton-Leibniz formula.