Example. Evaluate the definite integral

$$
\int_{0}^{2} \frac{4 \sqrt{2-x}-\sqrt{2+x}}{(\sqrt{x+2}+4 \sqrt{2-x})(x+2)^{2}} d x
$$
## 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 $\frac{a x+b}{c x+d}$. Therefore, we transform the integrand, highlighting $\sqrt{\frac{2-x}{2+x}}:$

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

2. We apply the substitution $\frac{2-x}{2+x}=t^{2}$ :

$$
x=\frac{4}{t^{2}+1}-2, \quad d x=-\frac{8 t}{\left(t^{2}+1\right)^{2}} d t, \quad t(0)=1, \quad t(2)=0
$$

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

$$
\int_{0}^{2} \frac{4 \sqrt{\frac{2-x}{2+x}}-1}{\left(1+4 \sqrt{\frac{2-x}{2+x}}\right)(x+2)^{2}} d x=\int_{1}^{0} \frac{4 t-1}{4 t+1} \frac{\left(t^{2}+1\right)^{2}}{16}\left(-\frac{8 t}{\left(t^{2}+1\right)^{2}}\right) d t
$$

We compute the antiderivative of the rational function and apply the Newton-Leibniz formula:

$$
\frac{1}{2} \int_{0}^{1} \frac{t(4 t-1)}{4 t+1} d t=\left.\frac{1}{4} t^{2}\right|_{0} ^{1}-\left.\frac{1}{4} t\right|_{0} ^{1}+\left.\frac{1}{16} \ln |4 t+1|\right|_{0} ^{1}=\frac{\ln 5}{16}
$$

Answer. $\int_{0}^{2} \frac{4 \sqrt{2-x}-\sqrt{2+x}}{(\sqrt{x+2}+4 \sqrt{2-x})(x+2)^{2}} d x=\frac{\ln 5}{16}$.

Conditions of the Problems. Calculate the definite integrals.

1. $\int_{4}^{9} \frac{\sqrt{x}}{\sqrt{x}-1} d x$.
2. $\int_{0}^{2} \frac{1}{\sqrt{x+1}+\sqrt{(x+1)^{3}}} d x$.
3. $\int_{0}^{4} \frac{1}{1+\sqrt{2 x+1}} d x$.
4. $\int_{0}^{1} \sqrt[3]{\frac{2-x}{2+x}} d x$
5. $\int_{1}^{2} \frac{\sqrt{x-1}-\sqrt{x+1}}{\sqrt{x-1}+\sqrt{x+1}} d x$.
6. $\int_{1}^{8} \frac{x+\sqrt[3]{x^{2}}+\sqrt[6]{x}}{x(1+\sqrt[3]{x}} d x$
7. $\int_{1}^{2} x \sqrt{\frac{x-1}{x+1}} d x$. 8. $\int_{0}^{63} \frac{6-\sqrt{x+1}+\sqrt[4]{x+1}}{\sqrt{(x+1)^{3}}-7(x+1)-6 \sqrt[4]{(x+1)^{3}}} d x$.
8. $\int_{3}^{29} \frac{\sqrt[3]{(x-2)^{2}}}{3+\sqrt[3]{(x-2)^{2}}} d x . \quad$ 10. $\int_{1}^{2} \frac{4 \sqrt{2-x}-\sqrt{3 x-2}}{(\sqrt{3 x-2}+4 \sqrt{2-x})(3 x-2)^{2}} d x$.

Answers. $\quad 1.7+2 \ln 2 . \quad 2 . \pi / 6 . \quad 3.2-\ln 2 . \quad 4 . \quad 3(\sqrt[3]{9}-1) / 8$.
5. $[\sqrt{2}-1-\ln (1+\sqrt{2})] / 2 . \quad 6.9 / 2-3 \pi / 2+6 \operatorname{arctg} 2 . \quad 7 . \ln (2+\sqrt{3}) / 2$.
8. $\ln (16 / 81)$.
9. $8+3 \sqrt{3} 2 \pi / 2$
10. $\ln 5 / 16$.

## 8.6. Integration of Expressions

$$
R\left(x,\left(a^{2} \pm x^{2}\right)^{1 / 2}\right) \text { and } R\left(x,\left(x^{2}-a^{2}\right)^{1 / 2}\right)
$$

Problem Statement. Calculate definite integrals of the form:

a) $\int_{\alpha}^{\beta} R\left(x, \sqrt{a^{2}-x^{2}}\right) d x$;

b) $\int_{\alpha}^{\beta} R\left(x, \sqrt{a^{2}+x^{2}}\right) d x$;

c) $\int_{\alpha}^{\beta} R\left(x, \sqrt{x^{2}-a^{2}}\right) d x$;

where $R$ is a rational function.

Plan of Solution.

1. To eliminate the radical, we 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. Applying the formula for the change of variable in a definite integral, we obtain integrals of the form

$$
\int_{t_{1}}^{t_{2}} R_{1}(\sin t, \cos t) d t \quad \text { or } \quad \int_{t_{1}}^{t_{2}} R_{1}(\operatorname{sh} t, \operatorname{ch} t) d t
$$

3. We compute the obtained integrals using known substitutions or the method of reducing the degree.