Question

Ввычислите интеграл, пожалуйста :)​

Answer 1

Ответ:

$\int\limits \frac{dx}{ \sqrt[3]{ {(x + 6)}^{2} } - 2 \sqrt{x + 6} } = \\ = \int\limits\frac{dx}{ {(x + 6)}^{ \frac{2}{3} } - 2 {(x + 6)}^{ \frac{1}{2} } } \\ \\ {(x + 6)}^{ \frac{1}{6} } = t \\ x + 6 = {t}^{6} \\ {(x + 6)}^{ \frac{1}{3} } = {t}^{2} \\ {(x + 6)}^{ \frac{2}{3} } = {t}^{4} \\ {(x + 6)}^{ \frac{1}{2} } = {t}^{3} \\ \\ dx = 6 {t}^{5} dt \\ \\ \int\limits \frac{6 {t}^{5}dt }{ {t}^{4} - 2 {t}^{3} } = 6\int\limits \frac{ {t}^{5}dt }{t {}^{3}( t - 2)} = 6\int\limits \frac{t {}^{2} dx}{t - 2} = \\ = 6\int\limits(t + 2 + \frac{4}{t - 2} )dt = \\ = 6( \frac{ {t}^{2} }{2} + 2t + 4 ln(t - 2)) + C = \\ = 3 {t}^{2} + + 12t + 24 ln(t - 2) + C = \\ = 3 \sqrt[3]{x + 6} + 12 \sqrt[6]{x + 6} + 24 ln( \sqrt[6]{x + 6} - 2) + C$