Question

Помогите плз Номер 15(а,б,в), номер 16(а,б), номер 10( 1 столбик), номер 11 (1столбик).

Answer 1

$15)\; \; a)\; \int (3p-1)^4\, dp=\frac{1}{3}\cdot \frac{(3p-1)^5}{5}+C\\\\\int \frac{dx}{(7x+5)^3}=\int (7x+5)^{-3}\, dx=\frac{1}{7}\cdot \frac{(7x+5)^{-2}}{-2}+C=-\frac{1}{14(7x+5)^2}+C\\\\\int \sqrt{2-3x}\, dx=\int (2-3x)^{1/2}\, dx=\frac{(2-3x)^{3/2}}{3/2}+C=\frac{2}{3}\cdot \sqrt{(2-3x)^3}+C\\\\16a)\; \; \int cos3x\, dx=\frac{1}{3}sin3x+C\\\\16b)\; \; \int sin\frac{x}{7}\, dx= -7cos\frac{x}{7}+C$

$10a)\; \; \int (3x+5)\, dx=\frac{1}{3}\cdot \frac{(3x+5)^2}{2}+C\\\\ili\; \; \int (3x+5)dx=3\cdot \frac{x^2}{2}+5x+C\\\\11a)\; \; \int (\frac{1}{x^2}-\sqrt{x})dx=\int (x^{-2}-\sqrt{x})dx=\frac{x^{-1}}{-1}-\frac{1}{2\sqrt{x}}+C=\\\\=-\frac{1}{x}-\frac{1}{2\sqrt{x}}+C\\\\11b)\; \; \int (\frac{3}{\sqrt{x}}-x^3)dx=\int (3x^{-1/2}-x^3)dx=3\cdot \frac{x^{\frac{1}{2}}}{1/2}-\frac{x^4}{4}+C=\\\\=6\sqrt{x}-\frac{x^4}{4}+C$