Example. Find the indefinite integral

$$
\int \frac{4 \sqrt{2-x}-\sqrt{2+x}}{(\sqrt{x+2}+4 \sqrt{2-x})(x+2)^{2}} d x
$$
## Solution.

1. To make a substitution leading to an integral of a rational function, it is necessary to transform the integrand so that it contains roots of any degree, but from the same expression of the form $\frac{a x+b}{c x+d}$.

Let's transform the integrand, highlighting $\sqrt{\frac{2-x}{2+x}}$:

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

2. Apply the substitution $t=\sqrt{\frac{2-x}{2+x}}$:

$$
x=\frac{4}{t^{2}+1}-2, \quad d x=-\frac{8 t}{\left(t^{2}+1\right)^{2}} d t
$$

By making the variable substitution in the indefinite integral, we get

$$
\int \frac{4 \sqrt{\frac{2-x}{2+x}}-1}{\left(1+4 \sqrt{\frac{2-x}{2+x}}\right)(x+2)^{2}} d x=\int \frac{4 t-1}{4 t+1} \frac{\left(t^{2}+1\right)^{2}}{16}\left(-\frac{8 t}{\left(t^{2}+1\right)^{2}}\right) d t
$$

3. Compute the antiderivative of the rational function $t$:

$$
\begin{aligned}
\frac{1}{2} \int \frac{t(4 t-1)}{4 t+1} d t=\frac{1}{2}\left[\int\left(t-\frac{1}{2}\right)\right. & \left.d t+\frac{1}{2} \int \frac{d t}{4 t+1}\right]= \\
& =\frac{1}{4} t^{2}-\frac{1}{4} t+\frac{1}{16} \ln |4 t+1|+C
\end{aligned}
$$

Return to the variable $x$ by substituting $t=\sqrt{\frac{2-x}{2+x}}$.

Answer. $\int \frac{4 \sqrt{2-x}-\sqrt{2+x}}{(\sqrt{x+2}+4 \sqrt{2-x})(x+2)^{2}} d x=$

$$
=\frac{1}{4} \frac{2-x}{2+x}-\frac{1}{4} \sqrt{\frac{2-x}{2+x}}+\frac{1}{16} \ln \left(4 \sqrt{\frac{2-x}{2+x}}+1\right)+C
$$

Conditions of the Problems. Find the indefinite integrals.

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

Answers.

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

## 7.9. Integration of expressions $R\left(x, \sqrt{a^{2} \pm x^{2}}\right)$ and $R\left(x, \sqrt{x^{2}-a^{2}}\right)$

Problem Statement. Find the indefinite integrals of the form:

a) $\int R\left(x, \sqrt{a^{2}-x^{2}}\right) d x$

b) $\int R\left(x, \sqrt{a^{2}+x^{2}}\right) d x$

c) $\int R\left(x, \sqrt{x^{2}-a^{2}}\right) d x$

where $R$ is a rational function.

Plan of Solution.

1. To eliminate the radical, use trigonometric or hyperbolic substitutions:
a) $x=a \sin t$ or $x=a$ th $t$;

b) $x=a \operatorname{tg} t$ or $x=a \operatorname{sh} t$;

c) $x=\frac{a}{\cos t}$ or $x=a \operatorname{ch} t$.

2. By applying the formula for the change of variable in an indefinite integral, we obtain integrals of the form

$$
\int R_{1}(\sin t, \cos t) d t
$$

3. Compute the last integral using known substitutions or the method of reducing the degree.
4. Return to the variable $x$ and write the answer.