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

1 Answer
May 22, 2018

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