∫ Integrals by algebraic substitution?

Question

can't get this ones!!! I think the answers are 1.45x10^39 0.445 and 17.125..... but why_ please help!!

∫ [0 to 1 ] (1-√x)^50 dx


∫ [0 to 4] dx/((1+√x)^3)

∫ [1 to 64] cubit root of X / ( 2 + cubic root of (X^2)) dx

Answer 1

(1) Let y = √x, so y^2 = x
∫ [0 to 1 ] (1-√x)^50 dx
= ∫ [0 to 1 ] (1 - y)^50 2ydy
= ∫ [0 to 1 ] -2y/51 d[(1 - y)^51]
= [-2y/51 (1 - y)^51][evaluate 0 to 1] + ∫ [0 to 1 ] 2/51 [(1 - y)^51] dy
= ∫ [0 to 1 ] -2/(51*52) d[(1 - y)^52]
= [-1/(51*26) (1 - y)^52] [evaluate 0 to 1]
= 1/1326
(2) Also let y = √x, so y^2 = x
∫ [0 to 4] dx/((1+√x)^3)
= ∫ [0 to 2] 2ydy / (1+ y)^3
= ∫ [0 to 2] [2(1 + y) - 2]dy / (1+ y)^3
= ∫ [0 to 2] [2/(1 + y)^2 - 2/ (1+ y)^3]dy
= [-2/(1 + y) + 1/ (1+ y)^2][evaluate 0 to 2]
(3) Let y = cubit root of X, and then you can solve it easily

Answer 2

enable (x+2) = t, then dx = dt, and 2x+a million = 2t - 3 J = ?(2x+a million)dx/(x+2)^2/3 = ?(2t-3)dx/t^(2/3) = 2?t^(a million/3)dt -3?t^(-2/3)dt = 3/2*t(t-6)/t^(2/3) + C ~~~~ ok, now purely replace t with the aid of applying x + 2, you will get: 3/2*t(t-6)/t^(2/3) + C = -9t^(a million/3) + 3/2*t^(4/3) + C = 3/2*(x+2)^4/3 - 9*(x+2)^a million/3 +C