Motivations behind substitutions in the integrals $\int\frac{\mathrm dx}{(ax+b)\sqrt{px^2+qx+r}}$ and $\int\frac{\mathrm dx}{(ax^2+b)\sqrt{cx^2+d}}$

Question

I was reading a textbook on integration where I came across suggested substitutions for certain types of integrals. These were as follows:

For evaluating integrals of the form $$\int\frac{\mathrm dx}{(ax+b)\sqrt{px^2+qx+r}}$$ Put $ax+b=\dfrac 1t$

For evaluating integrals of the form $$\int\frac{\mathrm dx}{(ax^2+b)\sqrt{cx^2+d}}$$ Put $x=\dfrac 1t$

I cannot understand the intuition/motivation behind these substitutions. I know that these work (I had tried them with a few examples). However, these substitutions would never strike me logically.

Typically, we use substitutions which would simplify the integral. However, until actually transforming the integrand, I cannot understand why these substitutions are useful; thus, they do not feel intuitive to me and would never strike me ad hoc.

Could somebody please explain to me the logic behind using these substitutions? As in without first modifying the integrand and seeing whether the modified integral is easier, how could it be understood that these substitutions would actually work and simplify the integrand?

P.S. Perhaps the integrals above are meant to be converted into the following integrals (whose closed forms are known to me):

$\displaystyle\int \frac{\mathrm dx}{x^2\pm a^2},\displaystyle\int \frac{\mathrm dx}{\sqrt{x^2\pm a^2}},\displaystyle\int \frac{\mathrm dx}{\sqrt{a^2-x^2}},\displaystyle\int\sqrt{x^2\pm a^2}\,\mathrm dx,\displaystyle\int\sqrt{a^2-x^2}\,\mathrm dx$

Answer 1

$\textbf{1.}$ First, let’s consider the specific case of $a=1, b=0$ to develop some intuition on how to integrate the general form.

So, we have $\displaystyle\int\frac{\mathrm dx}{x\sqrt{px^2+qx+r}}$. A standard technique to integrate this is by taking $x^2$ common from the square root, we get $\text{sgn}(x)\displaystyle\int\frac{\mathrm dx}{x^2\sqrt{\frac{r}{x^2}+\frac{q}x+p}}$. This can be easily integrated by substituting $t=\frac1x$ as its derivative is directly available in the integrand and the expression in the square root is a quadratic in $\frac1x$.

Now, this technique can be generalized by replacing $x$ with a linear expression in $x$, say $L=ax+b:$

$$\int\frac{\mathrm dx}{L\sqrt{px^2+qx+r}}=\frac1a\int\frac{\mathrm dL}{L\sqrt{PL^2+QL+R}}$$

This is why integrals of this form can be evaluated by substituting for the reciprocal of the linear expression.

$\textbf{2.}$ Notice that we have $\sqrt{cx^2+d}$ in the denominator. This is reminiscent of the derivative of $\sqrt{cx^2+d}$, just that we don’t have $x$ in the numerator. So, why not find a way to introduce $x$ in the numerator?

Observe that the effective power of $x$ in the integrand, i.e., the difference of the greatest powers of $x$ that can be taken common from the numerator and the denominator of the integrand, is $-3$ because from the denominator, $x^2$ can be taken common from $ax^2+b$ and $x$ from $cx^2+d$ and the numerator is $x^0$. We wish to make it $-2$ since we want it to be of the form $\dfrac{x}{(Ax^2+B)\sqrt{Cx^2+D}}$. So, let’s try substituting $x=t^n$ and see if there is a value of $n$ that makes the effective power of the new variable, $t$, as $-2$. We get:

$$\int\frac{\mathrm dx}{(ax^2+b)\sqrt{cx^2+d}}=n\int\frac{t^{n-1}\mathrm dt}{(at^{2n}+b)\sqrt{ct^{2n}+d}}$$

The effective power of $t$ is $-2n-1$ if $n>0$ and $n-1$ if $n<0$. We want it to be $-2$. Note that for the first condition, we obtain $n=-\frac12$ which is not possible. On considering the condition for $n<0$, we see that it is satisfied by $n=-1$. This explains why substituting $x=\frac1t$ in integrals of this type simplifies the integration process.