Question

Интеграл

( arctg^(37/60) (x^4+5) ) / (x^5 + 10x + 26/x^3 )

Answer 1

$\int {\frac{\sqrt[60]{arctg^{37} \, (x^4+5) }}{x^5 +10x +\frac{26}{x^3}}} \, dx = \int {\frac{x^3 \cdot arctg^{\frac{37}{60}} \, (x^4+5) }{x^8 +10x^4 +26}} \, dx=(*) \\ \\ t=arctg(x^4+5); \ \ dt = \frac{4x^3 \, dx}{1+x^8+10x+25}; \ \ dx= \frac{x^8 +10x+26 }{4x^3}\, dt \\ \\ (*) = \int {\frac{x^3 \cdot t^\frac{37}{60}}{x^8 +10x+26 } \cdot \frac{x^8 +10x+26 }{4x^3}\, dt =\frac{1}{4} \int {t^\frac{37}{60}} \, dt=\frac{1}{4 } \cdot \frac{60}{97} \cdot t^\frac{97}{60}+C}=$


$\\ \\ = \frac{15}{97} \cdot arctg^\frac{97}{60} \, (x^4+5)+C$