• 2022-05-26
    求积分∫上限2下限0{∣x-1∣/(x+1)(x-3)}dx
  • 先化简(x-1)/(x+1)(x-3)=[1/(x+1)(x-3)]*(x-1)=[1/(x-3)-1/(x+1)]*(x-1)/4=[(x-1)/(x-3)-(x-1)/(x+3)]/4={1+2/(x-3)-[1-2/(x+1)]}/4=[1/(x-3)+1/(x+1)]/2∫上限2下限0{∣x-1∣/(x+1)(x-3)}dx=-∫上限1下限0{(x-1)/(x+1)(x-3)}dx+∫上限2下限1{(x-1)/(x+1)(x-3)}dx=∫上限2下限1{[1/(x-3)+1/(x+1)]/2}dx-∫上限1下限0{[1/(x-3)+1/(x+1)]/2}dx=[ln|2-3|+ln|2+1|]-[ln|1-3|+ln|1+1|]-{[ln|1-3|+ln|1+1|]-[ln|0-3|+ln|0+1|]}=0+ln3-ln2-ln2-ln2-ln2+ln3+0=2ln3-4ln2

    内容

    • 0

      下列四个积分中,()是广义积分。 A: \( \int_0^2 { { 1 \over { { {(3 - x)}^2}}}dx} \) B: \( \int_0^6 { { {(x - 4)}^{ - {2 \over 3}}}dx} \) C: \( \int_0^1 { { 1 \over {1 + {x^2}}}dx} \) D: \( \int_1^2 { { 1 \over { { x^2}}}dx} \)

    • 1

      f(x)=x(x+1)(x+2(x-3), f'(0)=( ) A: 0 B: 1 C: -3 D: -6

    • 2

      求定积分:∫(上限ln2,下限0)(e^x)(1+e^x)dx的值.

    • 3

      ∫(上限л/2下限0)sin^2(x/2)dx

    • 4

      已知f(x)=x(x-1)(x-2)(x-3),则f'(0)=( ) A: 0 B: 3! C: -3! D: 1