∫ 1/x^2(1-x)dx Seeking indefinite integral

∫ 1/x^2(1-x)dx Seeking indefinite integral

The answer: 1 / [(1-x) x (1-x) x ^ 2] = A / (1-x (1-x) x (1-x) x (1-x) x ((1-x) x (2-bx + c-bx-2-cx) / [(1-x) x (1-x) x ((1-x) x ^ 2) / [(1-x) x (1-x) x (1-x) x (x + x) + (x + X + X + X (2) and (x + 2) as the solution is: a = b = b = C = 1 {{{[1 / [(1-x) x (1 (1 / (1-x) + [1 (1-x (1-x) (1-x (1-x-x-x) / X (1) / X (x (1) / x ^ 2) as (x ^ 2) is the [x (x (x (x + 1) / x ^ 2] as the [x ^ x + C

Detailed steps of ∫ (0 → 4) 1 / (x∧2-x-2) DX

1 / (x ^ 2-x-2) = 1 / [(X-2) (x + 1)} = (1 / 3) [1 / (X-2) - 1 / (x + 1)] = (1 / 3) [ln (X-2) - ln (x + 1) 'so ∫ (0 → 4) 1 / (x ∧ 2-x-2) DX = ∫ (0 → 4) d [ln (X-2) / 3-ln (x + 1) / 3] = LN2 / 3-ln5 / 3 - [ln (- 2) / 3-ln1 / 3] seems a little wrong? The given range of X is not right

∫ √ X / (2 - √ x) DX (the upper and lower limits of the integral are 0 to 1)

Let a = 2 - √ XX = (2-A) &# 178; DX = (2a-4) Da, so the original formula = ∫ (2,1) (2-A) / A * (2a-4) Da = - 2 ∫ (2,1) (A & # 178; - 4A + 4) / ADA = - 2 ∫ (2,1) (A-4 + 4 / a) Da = - A & # 178; + 8a-4lna (2,1) = (- 1 + 8-0) - (- 4 + 16-4ln2) = 4ln2-5