Question

Пожалуйста. Найти интеграл

Answer 1

=4∫√(1-x²)/(x²-1) dx +∫(3x²-3+3)/(x²-1) dx = -4∫1/√(1-x²) dx + 3∫(x²-1)/(x²-1) dx +
∫3/(x²-1) dx=-4arcsin(x) +3∫dx +(3/2)(∫1/(x-1)dx - ∫1/(x+1) dx) =
 -4arcsin(x) +3x + (3/2)(∫1/(x-1) d(x-1) - ∫1/(x+1) d(x+1)=
-4arcsin(x) +3x +(3/2)(ln(x-1)-ln(x+1) +C =
-4arcsin(x) +3x +(3/2)ln((x-1)/(x+1)) + C

Answer 2

можно на листочки. Фото. Если не трудно