Calculating $\int\frac{1}{(x^2+2x+3)\sqrt{x}}\,dx$

Question

My attempt:

Factor the denominator: The denominator can be factored as $(x + 1)(x + 3)$.

Substitution: Let's substitute $u = x + 1$. Notice that $du = dx$.

Rewrite the integral: Substitute $x$ and $dx$ in terms of $u: \displaystyle\int\dfrac{1}{(x+1)(x+3)\sqrt{x}}~dx$ becomes $\displaystyle\int\dfrac{1}{u(u+2)(\sqrt{u-1})}~du$.

I could go further. Please help me.

Answer 1

Your factorization of the quadratic in the denominator is incorrect: In fact, the quadratic is irreducible over $\Bbb R$, since its discriminant is negative: $2^2 - 4 \cdot 1 \cdot 3 = -8$.

Hint Changing variables via $u = \sqrt x$ rationalizes the integral, so that it can be handled in the usual way: $$\int \frac{du}{u^4 + 2 u^2 + 3} .$$ Now see this problem, which is identical except for the constants.

Answer 2

Let $a_\pm =\sqrt{2(\sqrt3\pm 1)}$ \begin{align} &\int\frac{1}{(x^2+2x+3)\sqrt{x}}\,dx =\int\frac{\frac1{\sqrt{x^3}}}{x+\frac3x+2}\ dx\\ =&\ \frac1{\sqrt3}\int\frac{(\frac1{2\sqrt{x}}+\frac{\sqrt3}{2\sqrt{x^3}})-( \frac1{2\sqrt{x}}-\frac{\sqrt3}{2\sqrt{x^3}})}{x+\frac3x+2}\ dx\\ =&\ \frac1{\sqrt3}\int\frac{d(\sqrt{x}-\frac{\sqrt3}{\sqrt{x}})}{a_+^2+(\sqrt{x}-\frac{\sqrt3}{\sqrt{x}}) ^2} + \frac{d(\sqrt{x}+\frac{\sqrt3}{\sqrt{x}})}{a_-^2-(\sqrt{x}+\frac{\sqrt3}{\sqrt{x}}) ^2} \\ =& \ \frac1{\sqrt3 a_+}\tan^{-1}\frac{{x}-{\sqrt3}}{a_+\sqrt x} + \frac1{\sqrt3 a_-}\coth^{-1}\frac{{x}+{\sqrt3}}{a_-\sqrt x} \end{align}