Question

Помогите пожалуйста!!! Интегралы

Answer 1

1) 1/4arctg(x/4)-c
2) arcsin(x/5)+c
3)1/12arctg(3x/4)+c

Answer 2

$\it\displaystyle\int\frac{dx}{x^2+16}=4\int\frac{d(\frac{x}{4})}{16(\frac{x^2}{16}+1)}=\frac{1}{4}arctg\frac{x}{4}+C\\\\\int\frac{dx}{\sqrt{25-x^2}}= 5\int\frac{d(\frac{x}{5})}{5\sqrt{1-\frac{x^2}{25}}}=arcsin\frac{x}{5}+C\\\\\int\frac{dx}{9x^2+16}=\frac{4}{3}\int\frac{d(\frac{3x}{4})}{16(\frac{9x^2}{16}+1)}=\frac{1}{12}arctg\frac{3x}{4}+C$