Question

ИНТЕГРАЛЫ, помогите, пожалуйста

Answer 1

Замена
x⁴ = t
4x³dx = dt
x³dx = dt/4
x⁸ = t²

$\int\limits { \frac{x^3dx}{x^8-2} } \, = \int\limits { \frac{dt}{4(t^2-2)} } \, = \frac{1}{4} \int\limits { \frac{dt}{t^2-2} } \, = \frac{1}{4} * \frac{1}{2 \sqrt{2} } *ln| \frac{t-\sqrt{2} }{t+\sqrt{2} } |+c= \\ \\ = \frac{1}{8 \sqrt{2} } ln| \frac{t-\sqrt{2} }{t+\sqrt{2} } |+c= \frac{1}{8 \sqrt{2} } ln| \frac{x^4-\sqrt{2} }{x^4+\sqrt{2} } |+c=$

Answer 2

$\int\limits { \frac{x^3}{x^8-2} } \, dx = \int\limits { \frac{1}{x^8-2} } \, d( \frac{x^4}{4}) = \frac{1}{4} \int\limits { \frac{1}{(x^4)^2-(\sqrt{2})^2} } \, d(x^4) =\\ \\ = \frac{1}{4} * \frac{1}{2\sqrt{2} } ln| \frac{x^4- \sqrt{2} }{x^4+ \sqrt{2}}| +C= \frac{1}{8 \sqrt{2} } ln| \frac{x^4- \sqrt{2} }{x^4+ \sqrt{2}}| +C$