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int(4e^x+6e^(-x))/(9e^x-4e^(-x))dx=A x+B...

`int(4e^x+6e^(-x))/(9e^x-4e^(-x))dx=A x+Blog(9e^(2x)-4)+C ,t h e n` A=_________ _, B=_________, C=___________

Text Solution

Verified by Experts

The correct Answer is:
`A=-3/2,B =35/36` C = any real constant
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Knowledge Check

  • If int(2e^(x)+3e^(-x))/(3e^(x)+4e^(-x))dx=Ax+B log(3e^(2x)+4) , then

    A
    `A=-(3)/(4), B=(1)/(24)`
    B
    `A=(3)/(4), B=-(1)/(24)`
    C
    `A=(1)/(4), B=(1)/(24)`
    D
    `A=-(3)/(4), B=(1)/(4)`

  • int(2e^(x))/(3+4e^(x))dx=

    A
    none of the following
    B
    `0.25log(3+4e^(x))+c`
    C
    `0.50log(3+4e^(x))+c`
    D
    `0.75log(3+4e^(x))+c`

  • If int(3e^(x)+4e^(-x))/(5e^(x)+6e^(-x))=ax+b log(5e^(2x)+6)+c, then

    A
    `a=(2)/(3), b=-30`
    B
    `a=(3)/(2), b=(-1)/(30)`
    C
    `a=(2)/(3), b=(-1)/(30)`
    D
    `a=-(3)/(2), b=-30`

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