Evaluating $\int\frac{\mathrm dx}{Q_1\sqrt Q_2}$ by $x=\frac1{t}$ When the Coefficients of $x$ in $Q_1$ and $Q_2$ are Non-Zero

Question

While teaching about the general technique to integrate irrational functions$\left(\text{like} \int\frac{\mathrm dx}{L_1\sqrt L_2}, \int\frac{\mathrm dx}{L\sqrt Q}, \int\frac{\mathrm dx}{Q\sqrt L}\text{ and }\int\frac{\mathrm dx}{Q_1\sqrt Q_2}\right)$, my teacher taught that integrals of the form $\int\frac{\mathrm dx}{Q_1\sqrt Q_2}$ where $Q_1$ and $Q_2$ are quadratic functions in $x$ can be integrated by the substitution $x=\frac1{t}$. This works well in examples where the coefficient of $x$ in both the quadratics is zero as I can take the resulting quadratic in the square root as $u$, but when it's not, I am not able to proceed with this substitution as it seems to complicate the integral. Can someone please explain how to do so?

$\big{(}$I am aware of hyperbolic substitutions to eliminate $\sqrt Q_2$, but here I am specifically interested in the substitution $x=\frac1{t}$.$\big{)}$

For example,

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

Putting $x=\frac1{t}$, $$\mathcal I=-\text{sgn}x\int\frac{t\mathrm dt}{(t^2+t+1)\sqrt{3t^2+2t+1}}$$ $$=-\frac{\text{sgn}x}{\sqrt3}\int\frac{t\mathrm dt}{(t^2+t+1)\sqrt{\left(t+\frac13\right)^2+\frac29}}$$

Then, I tried to rewrite the numerator as $t=\frac12\left(2t+\frac23\right)-\frac13$ and take $t^2+\frac{2t}3+\frac13=u^2$ for the first integral:

$$\mathcal I=-\frac{\text{sgn}x}{\sqrt3}\int\frac{\mathrm du}{u^2+\frac13\sqrt{u^2-\frac29}+\frac59}-\frac{\text{sgn}x}{3\sqrt3}\int\frac{\mathrm dt}{(t^2+t+1)\sqrt{t^2+\frac{2t}3+\frac13}}$$

For the first integral, I thought of rationalizing the denominator, but it seems to complicate it even more. And ironically, the second integral is itself of the very form I wish to compute.

Answer 1

Not really a complete answer, but it may help:

When the coefficients of $x$ and $x^2$ in $Q_{1}$ and $Q_{2}$ have the same (non zero) ratio, you can complete the square, and find a suitable substitution.

For example,

$Q_{1}\sqrt{Q_{2}} \ := \ (2x^2+8x+9)\sqrt{x^2+4x+7} \ = \ \ (2(x+2)^2+1)\sqrt{(x+2)^2+3} \ $ and then the substitution $u=x+2 \ $ reduces the integrand to your required form.

If that doesn't work, we can try factoring $Q_{1}$ to get an integrand of the form $ \frac{1}{L_{1} L_{2}\sqrt{Q_2}}$ Where $L_{1}$ and $L_{2}$ are linear functions of $x$. From here, we proceed with partial fractions to get an integrand of the form $ \frac{a}{L_{1} \sqrt{Q_{2}}} + \frac{b}{L_{2} \sqrt{Q_{2}}} $ which can be solved using the substitution $L=\frac{1}{t}$.

In your particular example, this may not work either since $Q_{1}$ has complex roots, so you'll probably have to resort to trigonometric/hyperbolic substitutions.

Sorry if this wasn't particularly helpful.