If ∫2ex+3e−x3ex+4e−xdx=Ax+Blog⁡3e2x+4+C, then

If 2ex+3ex3ex+4exdx=Ax+Blog3e2x+4+C, then

  1. A

    A=34,B=124

  2. B

    A=34,B=124

  3. C

    A=14,B=124

  4. D

    A=34,B=14

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    Solution:

    Let

    2ex+3ex3ex+4exdx=2e2x+33e2x+4dx 2e2x+3=A3e2x+4+B6e2x 2e2x+3=(3A+6B)e2x+4A

    On comparing both sides, we get

    2=3A+6B----i3=4A----ii

    From Eqs. (i) and (ii), woe get

     2=94+6B 6B=294=14  B=124 2e2x+33e2x+4dx=33e2x+443e2x+4dx1246e2x3e2x+4dx =34x124log3e2x+4+C

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