2007-08-20 03:20:05 · 4 answers · asked by Gabriel 3 in Science & Mathematics ➔ Mathematics
I = ∫ (3x + 2)^( - 3 / 2 ) dx
Let u = 3x - 2
du = 3 dx
du / 3 = dx
I = (1 / 3) ∫ u ^(- 3 / 2) du
I = (1 / 3)(- 2) u^(- 1 / 2) + C
I = (- 2) / [ (3) (3x - 2)^(1 / 2) ] + C
2007-08-20 08:23:19 · answer #1 · answered by Como 7 · 1⤊ 1⤋
Integral (3x-2)^(-3/2) dx = 1/3 Int (u^(-3/2)) du dx =
1/3(u^-.5)/-.5 = -2/(3(3x-2)^.5) + C
2007-08-20 03:31:43 · answer #2 · answered by John V 6 · 1⤊ 0⤋
1 /â ((3x-2)^3)
= 1 / (3x-2)^(3/2)
= (3x-2)^-(3/2)
so you need to find
â« (3x-2)^-(3/2) dx
let u = 3x -2
so du/dx = 3 gives dx = du /3
then
â« (3x-2)^-(3/2) dx simplifies to
â« u^-(3/2) du/3
= (1/3) â« (u)^-(3/2) du
= (1/3) [ (u^-(1/2)) /-(1/2)]
= -(2/3) u^-(1/2) then resubstitute u =3x-2 gives
= -(2/3) (3x-2)^-(1/2)
= -2 /[3â(3x-2)] + C
the end
.
2007-08-20 03:30:09 · answer #3 · answered by The Wolf 6 · 1⤊ 0⤋
1/â(3x-2)^3 using u = 3x-2, then du = 3dx
1/3u^-3/2du
-1/6u^-1/2 = 1/6â(3x-2)
2007-08-20 03:31:23 · answer #4 · answered by chasrmck 6 · 0⤊ 1⤋