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the integral from 0 to 4 of
(x+1)/(16+x^2) dx

the answers are log(sqrt2) + pi/16

Could you please take me through how to get to that answer please, thank you very much!

  • Math -

    split (x+1)/(16+x^2) into x/(16+x^2) + 1/(16+x^2)

    the integral of x/(16+x^2) = (1/2) ln(16+x^2)
    ( I recognized the ln derivative pattern)

    for the integral of 1/(16+x^2) I went to my "archaic" integral talbles and go
    (1/4) tan^-1 (x/4)

    so the integral would be
    (1/2)ln(16+x^2) + (1/4)tan^-1 (x/4)

    so from 0 to 4 would give me
    ( (1/2)ln(32) + (1/4)tan^-1 (1) - ( (1/2) ln16 + (1/4)tan^-1 (0) )
    = ln √32 + (1/4)(π/4) - ln √16 - (1/4)(0)
    = ln (4√2) + π/16 - ln 4 - 0
    = ln4 + ln √2 + π/16 - ln4

    = ln √2 + π/16

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