How do you find the integral #intx/(sqrt(x^2+x+1))dx# ?

1 Answer
May 22, 2018

Answer:

Use the substitution #2x+1=sqrt3tantheta#.

Explanation:

Let

#I=intx/sqrt(x^2+x+1)dx#

Complete the square in the denominator:

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

Apply the substitution #2x+1=sqrt3tantheta#:

#I=int(sqrt3tantheta-1)/(sqrt3sectheta)(sqrt3/2sec^2thetad theta)#

Simplify:

#I=1/2int(sqrt3secthetatantheta-sectheta)d theta#

Integrate term by term:

#I=1/2{sqrt3sectheta-ln|sectheta+tantheta|}+C#

Reverse the substitution:

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