Question

Evaluate the following integrals: $$ \int \frac{x^{2}+2 x+3}{\sqrt{\left(x^{2}+x+1\right)}} d x $$

Short answer

** The result for the integral of the rational function is: $$ \int \frac{x^{2}+2 x+3}{\sqrt{\left(x^{2}+x+1\right)}} d x = \frac{1}{2}\left((x+\frac{1}{2})^{2} + \frac{3}{4} \right) + C $$

Step-by-step solution

Complete the square in the denominator

We have the integral: $$ \int \frac{x^{2}+2 x+3}{\sqrt{\left(x^{2}+x+1\right)}} d x $$ We can make the expression inside the square root easier if we complete the square. We have: $$ x^2 + x + 1 = \left(x+\frac{1}{2}\right)^2 + \frac{3}{4} $$

Apply substitution

Now let's substitute: $$ u = x + \frac{1}{2} $$ So, $$ du = dx $$ Thus, the integral becomes: $$ \int \frac{(u- \frac{1}{2})^{2}+2 (u- \frac{1}{2})+3}{\sqrt{u^2 + \frac{3}{4}}} du $$

Simplify the integrand

Now we can simplify the integral: $$ \int \frac{u^2}{\sqrt{u^2 + \frac{3}{4}}} du $$

Apply another substitution

Now we substitute: $$ v^2 = u^2 + \frac{3}{4} $$ Differentiating both sides gives: $$ 2v dv = 2u du $$ Then, dividing both sides by \(2u\) we get: $$ \frac{v}{u} dv = du $$ The integral becomes: $$ \int v dv $$

Evaluate the integral and back-substitute

Integrating with respect to v, we have: $$ \frac{1}{2}v^2 + C $$ Now we can substitute back for u and x. First, let's substitute for u: $$ \frac{1}{2}(u^2 + \frac{3}{4}) + C $$ Then for x: $$ \frac{1}{2}\left( (x+\frac{1}{2})^{2} + \frac{3}{4} \right) + C $$ So, the final result for the integral is: $$ \int \frac{x^{2}+2 x+3}{\sqrt{\left(x^{2}+x+1\right)}} d x = \frac{1}{2}\left((x+\frac{1}{2})^{2} + \frac{3}{4} \right) + C $$