Question

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

Answer 1

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`