Math
posted by Candice .
Evaluate
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!

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