Упр.22.23 ГДЗ Мордкович Семенов 11 класс (Алгебра)

1. Найдите неопределённый интеграл:
   а) $$\int\left(3\cdot 2^x-\frac{2}{2\cos^2\left(\frac{x}{2}\right)}+\sqrt{3x+1}\right)\,dx$$;
   б) $$\int\left(\frac{5}{1+x^2}-\frac{3}{\sqrt{1-x^2}}+2\sin 2x\right)\,dx$$;
   в) $$\int\left(2e^{3x-1}+\frac{5}{\sin^2 5x}-4(4x+1)^{\frac{1}{5}}\right)\,dx$$;
   г) $$\int\left(\cos^2 2x-\left(\frac{2x^5}{3x^7+3x^5}+\frac{1}{\sqrt{9-9x^2}}\right)\right)\,dx$$.

а)

$$\int \left(3\cdot 2^x-\frac{2}{2\cos^2 \frac{x}{2}}+\sqrt{3x+1}\right)\,dx = \int 3\cdot 2^x\,dx-\int \frac{1}{\cos^2 \frac{x}{2}}\,dx+\int \sqrt{3x+1}\,dx.$$

$$\int 3\cdot 2^x\,dx=\frac{3\cdot 2^x}{\ln 2},\qquad \int \frac{1}{\cos^2 \frac{x}{2}}\,dx=2\tan \frac{x}{2},$$
$$\int \sqrt{3x+1}\,dx=\int (3x+1)^{1/2}\,dx=\frac{2}{9}(3x+1)^{3/2}.$$

$$\int \left(3\cdot 2^x-\frac{2}{2\cos^2 \frac{x}{2}}+\sqrt{3x+1}\right)\,dx = \frac{3\cdot 2^x}{\ln 2}-2\tan \frac{x}{2}+\frac{2}{9}(3x+1)^{3/2}+C.$$

б)

$$\int \left(\frac{5}{1+x^2}-\frac{3}{\sqrt{1-x^2}}+2\sin 2x\right)\,dx = 5\int \frac{dx}{1+x^2}-3\int \frac{dx}{\sqrt{1-x^2}}+2\int \sin 2x\,dx.$$

$$5\int \frac{dx}{1+x^2}=5\arctan x,\qquad -3\int \frac{dx}{\sqrt{1-x^2}}=-3\arcsin x,$$
$$2\int \sin 2x\,dx=-\cos 2x.$$

$$\int \left(\frac{5}{1+x^2}-\frac{3}{\sqrt{1-x^2}}+2\sin 2x\right)\,dx = 5\arctan x-3\arcsin x-\cos 2x+C.$$

в)

$$\int \left(2e^{3x-1}+\frac{5}{\sin^2 5x}-4\sqrt[5]{4x+1}\right)\,dx = \int 2e^{3x-1}\,dx+5\int \csc^2 5x\,dx-4\int (4x+1)^{1/5}\,dx.$$

$$\int 2e^{3x-1}\,dx=\frac{2}{3}e^{3x-1},\qquad 5\int \csc^2 5x\,dx=-\cot 5x,$$
$$-4\int (4x+1)^{1/5}\,dx =-4\cdot \frac{1}{4}\cdot \frac{5}{6}(4x+1)^{6/5} =-\frac{5}{6}(4x+1)^{6/5}.$$

$$\int \left(2e^{3x-1}+\frac{5}{\sin^2 5x}-4\sqrt[5]{4x+1}\right)\,dx = \frac{2}{3}e^{3x-1}-\cot 5x-\frac{5}{6}(4x+1)^{6/5}+C.$$

г)

$$\int \left(\cos^2 2x-\frac{2x^5}{3x^7+3x^5}+\frac{1}{\sqrt{9-9x^2}}\right)\,dx.$$

Упростим дроби:
$$\frac{2x^5}{3x^7+3x^5}=\frac{2x^5}{3x^5(x^2+1)}=\frac{2}{3(x^2+1)}, \qquad \frac{1}{\sqrt{9-9x^2}}=\frac{1}{3\sqrt{1-x^2}}.$$
Также
$$\cos^2 2x=\frac{1+\cos 4x}{2}.$$

Тогда
$$\int \left(\cos^2 2x-\frac{2x^5}{3x^7+3x^5}+\frac{1}{\sqrt{9-9x^2}}\right)\,dx = \int \left(\frac{1+\cos 4x}{2}-\frac{2}{3(x^2+1)}+\frac{1}{3\sqrt{1-x^2}}\right)\,dx.$$

$$\int \frac{1}{2}\,dx=\frac{x}{2},\qquad \int \frac{\cos 4x}{2}\,dx=\frac{1}{8}\sin 4x,$$
$$-\frac{2}{3}\int \frac{dx}{x^2+1}=-\frac{2}{3}\arctan x,\qquad \frac{1}{3}\int \frac{dx}{\sqrt{1-x^2}}=\frac{1}{3}\arcsin x.$$

$$\int \left(\cos^2 2x-\frac{2x^5}{3x^7+3x^5}+\frac{1}{\sqrt{9-9x^2}}\right)\,dx = \frac{x}{2}+\frac{1}{8}\sin 4x-\frac{2}{3}\arctan x+\frac{1}{3}\arcsin x+C.$$

Ответ

$$\text{а) } \frac{3\cdot 2^x}{\ln 2}-2\tan \frac{x}{2}+\frac{2}{9}(3x+1)^{3/2}+C;$$
$$\text{б) } 5\arctan x-3\arcsin x-\cos 2x+C;$$
$$\text{в) } \frac{2}{3}e^{3x-1}-\cot 5x-\frac{5}{6}(4x+1)^{6/5}+C;$$
$$\text{г) } \frac{x}{2}+\frac{1}{8}\sin 4x-\frac{2}{3}\arctan x+\frac{1}{3}\arcsin x+C.$$