Question

Помогите решить, срочно

Answer 1

$\int(\frac{4}{3}x^3-\frac{3}{2}x^2+8)\, dx=\frac{4}{3}\cdot \frac{x^4}{4}-\frac{3}{2}\cdot \frac{x^3}{3}+8x+C=\frac{x^4}{3}-\frac{x^3}{2}+8x+C\; ;\\\\\\\int (5^{x}+e^{x}+2x)\, dx=\frac{5^{x}}{\ln5}+e^{x}+2\cdot \frac{x^2}{2}+C=\frac{5^{x}}{\ln5}+e^{x}+x^2+C\; ;\\\\\\\int\limits^{\frac{\pi}{6}}_0cos2x\, dx=\frac{1}{2}sin2x\Big |_0^{\frac{\pi }{6}}=\frac{1}{2}\cdot \frac{\sqrt3}{2}=\frac{\sqrt3}{4}$

$\int\limits^{\sqrt3}_1\frac{dx}{1+x^2}\, dx=arctgx\Big |_1^{\sqrt3}}=arctg\sqrt3-arctg1=\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi }{12}$