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March 29, 2015

March 29, 2015

Posted by **Candice** on Thursday, April 19, 2012 at 6:26am.

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 -
**Reiny**, Thursday, April 19, 2012 at 8:14amsplit (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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