Question

不定积分计算？

dx/（（3-ax)(5-bx)）不定积分怎么算
求详细步骤

Answer 1

当 a = b = 0 时，I = ∫dx/[(3-ax)(5-bx)] = (1/15)∫dx = x/15 + C ;
当 a = 0, b ≠ 0 时，I = ∫dx/[(3-ax)(5-bx)] = (1/3)∫dx/(5-bx) = -1/(3b)ln|5-bx| + C;
当 a ≠ 0, b = 0 时，I = ∫dx/[(3-ax)(5-bx)] = (1/5)∫dx/(3-ax) = -1/(5a)ln|3-ax| + C;
当 a ≠ 0, b ≠ 0 ，b = (5/3)a 时
∫dx/[(3-ax)(5-bx)] = ∫dx/[(5/3)(3-ax)^2] = [-3/(5a)]∫d(3-ax)/(3-ax)^2
= [3/(5a)]/(3-ax) + C;
当 a ≠ 0, b ≠ 0 ，b ≠ (5/3)a 时,
设 1/[(3-ax)(5-bx)] = p/(3-ax) + q/(5-bx) = [p(5-bx)+q(3-ax)]/[(3-ax)(5-bx)]
= [(5p+3q)-(bp+aq)x]/[(3-ax)(5-bx)]
则 5p+3q = 1， bp+aq = 0, 得 p = a/(5a-3b), q = -b/(5a-3b)
∫dx/[(3-ax)(5-bx)] = ∫{[a/(5a-3b)]/(3-ax) - [b/(5a-3b)]/(5-bx)]}dx
= [-1/(5a-3b)]ln|3-ax| + [1/(5a-3b)]ln|5-bx| + C
= [1/(5a-3b)]ln|(5-bx)/(3-ax)| + C

Answer 2

大概是图里这样

Answer 3