Упр.361 ГДЗ Колягин Ткачёва 11 класс (Алгебра)
- $$e^{3x}-\cos 2x$$
- $$e^{\frac{x}{3}}+\sin 3x$$
- $$\frac{2\sin x}{3}-5e^{2x}+\frac{1}{5}$$
- $$\frac{3\cos x}{7}+2e^{3x}-\frac{1}{2}$$
- $$\frac{\sqrt[5]{x}}{4}-5\cos(6x-1)$$
- $$\frac{\sqrt{x}}{5}+4\sin(4x+2)$$
- $$\frac{3}{\sqrt[3]{2x-1}}$$
- $$\frac{4}{\sqrt{3x+1}}-\frac{3}{2x-5}$$
Найдём первообразную по частям:
$$\int \left(e^{3x}-\cos 2x\right)\,dx=\frac{1}{3}e^{3x}-\frac{1}{2}\sin 2x+C.$$
$$\int \left(e^{x/3}+\sin 3x\right)\,dx=3e^{x/3}-\frac{1}{3}\cos 3x+C.$$
$$\int \left(2\sin \frac{x}{3}-5e^{2x+\frac15}\right)\,dx=-6\cos \frac{x}{3}-\frac{5}{2}e^{2x+\frac15}+C.$$
$$\int \left(3\cos \frac{x}{7}+2e^{3x-\frac12}\right)\,dx=21\sin \frac{x}{7}+\frac{2}{3}e^{3x-\frac12}+C.$$
$$\int \left(\sqrt[5]{\frac{x}{4}}-5\cos(6x-1)\right)\,dx=\frac{10}{3}\left(\frac{x}{4}\right)^{6/5}-\frac{5}{6}\sin(6x-1)+C.$$
$$\int \left(\sqrt{\frac{x}{5}}+4\sin(4x+2)\right)\,dx=\frac{10}{3}\left(\frac{x}{5}\right)^{3/2}-\cos(4x+2)+C.$$
$$\int \frac{3}{\sqrt[3]{2x-1}}\,dx=\frac{9}{4}(2x-1)^{2/3}+C.$$
$$\int \left(\frac{4}{\sqrt{3x+1}}-\frac{3}{2x-5}\right)\,dx=\frac{8}{3}\sqrt{3x+1}-\frac{3}{2}\ln|2x-5|+C.$$
Ответ
- $$F(x)=\frac{1}{3}e^{3x}-\frac{1}{2}\sin 2x+C$$
- $$F(x)=3e^{x/3}-\frac{1}{3}\cos 3x+C$$
- $$F(x)=-6\cos \frac{x}{3}-\frac{5}{2}e^{2x+\frac15}+C$$
- $$F(x)=21\sin \frac{x}{7}+\frac{2}{3}e^{3x-\frac12}+C$$
- $$F(x)=\frac{10}{3}\left(\frac{x}{4}\right)^{6/5}-\frac{5}{6}\sin(6x-1)+C$$
- $$F(x)=\frac{10}{3}\left(\frac{x}{5}\right)^{3/2}-\cos(4x+2)+C$$
- $$F(x)=\frac{9}{4}(2x-1)^{2/3}+C$$
- $$F(x)=\frac{8}{3}\sqrt{3x+1}-\frac{3}{2}\ln|2x-5|+C$$







