Question

Вычислить интегралы, 3 вариант все

Answer 1

1. $\int\limits {\sqrt[3]{(1+4x)^2} } \, dx = \frac{1}{4} \int\limits {\sqrt[3]{(1+4x)^2} } \, d(1+4x) = \frac{1}{4} \frac{(1+4x)^\frac{5}{3} }{\frac{5}{3} } = \frac{3\sqrt[3]{(1+4x)^5} }{20} + c$

2. $\int\limits {\frac{1}{(2+3x)^5} } \, dx = \frac{1}{3} \int\limits {\frac{1}{(2+3x)^5} } \, d(2+3x) = \frac{1}{3} \frac{(2+3x)^6}{6} = \frac{(2+3x)^6}{18} + c$

3. $\int\limits {e^{2x+1}} \, dx = \frac{1}{2} \int\limits {e^{2x+1}} \, d(2x+1) = \frac{1}{2} e^{2x+1} + c$

4. $\int\limits {cos\frac{4}{3}x } \, dx = \int\limits {cos\frac{4}{3} } \, d(\frac{4}{3}x) = \frac{3}{4} sin\frac{4}{3}x + c$

5. $\int\limits {\frac{2}{\sqrt{4-x^2} } } \, dx = 2\int\limits {\frac{1}{\sqrt{4-x^2} } } \, dx = 2arcsin\frac{x}{2} + c$

6. $\int\limits {\frac{x}{\sqrt{4-x^2} } } \, dx = -\frac{1}{2} \int\limits {\frac{1}{\sqrt{4-x^2} } } \, d(4-x^2) = -\frac{1}{2} \frac{(4-x^2)^\frac{1}{2} }{\frac{1}{2} } } = -\sqrt{4-x^2}$ + c[/tex]

7. $\int\limits {\frac{sin3x}{cos3x} } \, dx = -\frac{1}{3} \int\limits {\frac{1}{cos3x} } \, d(cos3x) = -\frac{1}{3} ln |cos3x| + c$