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

May 22, 2018

Use the substitution $2 x + 1 = \sqrt{3} \tan \theta$.

Explanation:

Let

$I = \int \frac{x}{\sqrt{{x}^{2} + x + 1}} \mathrm{dx}$

Complete the square in the denominator:

$I = \int \frac{2 x}{\sqrt{{\left(2 x + 1\right)}^{2} + 3}} \mathrm{dx}$

Apply the substitution $2 x + 1 = \sqrt{3} \tan \theta$:

$I = \int \frac{\sqrt{3} \tan \theta - 1}{\sqrt{3} \sec \theta} \left(\frac{\sqrt{3}}{2} {\sec}^{2} \theta d \theta\right)$

Simplify:

$I = \frac{1}{2} \int \left(\sqrt{3} \sec \theta \tan \theta - \sec \theta\right) d \theta$

Integrate term by term:

$I = \frac{1}{2} \left\{\sqrt{3} \sec \theta - \ln | \sec \theta + \tan \theta |\right\} + C$

Reverse the substitution:

$I = \frac{1}{2} \sqrt{{\left(2 x + 1\right)}^{2} + 3} - \frac{1}{2} \ln | 2 x + 1 + \sqrt{{\left(2 x + 1\right)}^{2} + 3} | + C$