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int(x+1)/(x^2+5x6)dx=...

`int(x+1)/(x^2+5x6)dx=`

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int(1)/(x^(2)-5x+6)dx

int 1/(x^2-5x+6) dx

Knowledge Check

  • int(x+1)/(x^2+5x+6)dx=

    A
    `log|((x+3)^2)/(x+2)|+c`
    B
    `log|(x+2)/((x+3)^2)|+c`
    C
    `2log|(x+3)/(x+2)|+c`
    D
    `2log|(x+2)/(x+3)|+c`
  • int(5x-1)/(3x^2+x+2)dx=

    A
    `(5)/(6)log|3x^2+x+2|-(11)/(3sqrt(23))tan^(-1)((6x+1)/(sqrt(23)))+c`
    B
    `(5)/(6)log|3x^2+x+2|+(11)/(3sqrt(23))tan^(-1)((6x+1)/(sqrt(23)))+c`
    C
    `(5)/(3)log|3x^2+x+2|-(11)/(3sqrt(23))tan^(-1)((6x+1)/(sqrt(23)))+c`
    D
    `(5)/(3)log|3x^2+x+2|+(11)/(3sqrt(23))tan^(-1)((6x+1)/(sqrt(23)))+c`
  • int(x^2)/(x^4+5x^2+6)dx=

    A
    `sqrt(3)tan^(-1)((x)/(sqrt(3)))-sqrt(2)tan^(-1)((x)/(sqrt(2)))+c`
    B
    `sqrt(3)tan^(-1)((x)/(sqrt(3)))+sqrt(2)tan^(-1)((x)/(sqrt(2)))+c`
    C
    `(1)/(sqrt(3))tan^(-1)((x)/(sqrt(3)))-(1)/(sqrt(2))tan^(-1)((x)/(sqrt(2)))+c`
    D
    `(1)/(sqrt(3))tan^(-1)((x)/(sqrt(3)))+(1)/(sqrt(2))tan^(-1)((x)/(sqrt(2)))+c`
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    int_(2)^(3)(1)/(x^(2)+5x+6)dx =