integral 3-x dx/√x2-6x-12

jawab

u = x² - 6x  - 12
du = 2x - 6 dx
du = -2(3 - x) dx
(3-x) dx = - 1/2 du

∫ (3-x) dx/ √(x² -6x  -12) dx

= -1/2  ∫u^(-1/2) du

= -1/2  [ 1/2 ( u)^(1/2) ]+ c

= 1/4 √ u + c

= 1/4 √( x² - 6x  -12) + c