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I need help with the following integral:

7. (6t + 2) / (t^2 + 2t + 17)

I understand that one needs to complete-the-square with the denominator and then use a substitution. I am confused as to how to go about the latter part. Any help is much appreciated.

3 Answers

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  • 1 decade ago
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    int [ (6t + 2) / (t^2 + 2t + 17) dt ]

    = int [ (6t + 2) / [ (t + 1)^2 + 16] dt ]

    Let u = t + 1

    du = dt

    = int [ (6u - 4) / (u^2 + 16) du ]

    = 6 * int [ u / (u^2 + 16) du ] - int [ 4 / (u^2 + 16) ]

    for the first integral: v = u^2 + 16; (1 / 2)dv = u du

    = 6 * (1 / 2) * int [ (1 / v) dv ] - 4* int [ 1 / (u^2 + 16) du ]

    = 3 ln abs ( v ) - 4*(1 / 4)arctan (u / 4) + C

    = 3 ln abs (u^2 + 16) - arctan (u / 4) + C

    = 3 ln abs [ (t + 1)^2 + 16 ] - arctan [ (t + 1) / 4 ] + C

  • Anonymous
    1 decade ago

    The denominator is (t + 1)^2 + 16

    (6t + 2) / (t^2 + 2t + 17) = ( 6t + 2 ) / ( (t + 1)^2 + 16 )

    ** t + 1 = 4u

    ( 6t + 2 ) / ( 16u^2 + 16 )

    ( 6t + 6 - 4 ) / ( 16u^2 + 16 )

    ( 6(t + 1) - 4 ) / ( 16u^2 + 16 )

    ( 6(4u) - 4 ) / ( 16u^2 + 16 )

    ( 24u - 4 ) / ( 16u^2 + 16 )

    4*( 6u - 1 ) / 16*( u^2 + 1 )

    (1/4) * ( 6u - 1 ) / ( u^2 + 1 )

    (1/4) * [ 6u / ( u^2 + 1 ) - 1 / ( u^2 + 1 ) ]

    Ok, we have ::

    INTEGRAL (1/4) * [ 6u / ( u^2 + 1 ) - 1 / ( u^2 + 1 ) ] dt

    ** dt/du = 4

    ** dt = 4*du

    INTEGRAL [ 6u / ( u^2 + 1 ) - 1 / ( u^2 + 1 ) ] du

  • 1 decade ago

    when u solve u get

    14(3t+1)/(t+5)(t-3)

    ur denominator is (t+1)^2-4

    then solve

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