How do you find the integral of #(6x+1)/(x^2+2x+3)#?

1 Answer
maganbhai P.
Mar 18, 2018

#I=3ln|x^2+2x+3|-5/sqrt(2)tan^-1((x+1)/sqrt(2))+c#

Explanation:

#I=int(6x+1)/(x^2+2x+3)dx#

#=int(6x+6)/(x^2+2x+3)dx-int5/(x^2+2x+3)dx#

#I=3int(2x+2)/(x^2+2x+3)dx-int5/(x^2+2x+1+2)dx#

#I=3int(d/(dx)(x^2+2x+3))/(x^2+2x+3)dx-int5/((x+1)^2+(sqrt(2))^2)dx#

#I=3ln|x^2+2x+3|-5*1/sqrt(2)tan^-1((x+1)/sqrt(2))+c#
#I=3ln|x^2+2x+3|-5/sqrt(2)tan^-1((x+1)/sqrt(2))+c#

Related questions
See all questions in Integrals of Rational Functions
Impact of this question
1719 views around the world
You can reuse this answer
Creative Commons License