I wish someone could help me derive this expression. ($K$ is a constant coefficient. $P_n(x)$ is a polynomial function of degree n.) $$ \int\frac{P_n(x)\mathrm{d}x}{\sqrt{ax^2+bx+c}} \equiv P_{n-1}(x) \cdot\sqrt{ax^2+bx+c} + K\cdot\int\frac{\mathrm{d}x}{\sqrt{ax^2+bx+c}}, (a\neq0) $$ After finding the derivatives of both sides it is easy to find the coefficients of the polynomial $P_{n-1}(x)$. Then we are left with this simple integral: $$\int\frac{\mathrm{d}x}{\sqrt{ax^2+bx+c}}$$
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1$\begingroup$ What are $P_n(x)$ and $P_{n - 1}(x)$? It looks like an incorrect integration by parts. $\endgroup$– LSpiceCommented Mar 29, 2021 at 10:17
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$\begingroup$ Those are polynomial functions of degree n and n-1. $\endgroup$– RengarJGCommented Mar 29, 2021 at 10:19
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2$\begingroup$ try $a=0$, $b=0$, $c=1$, and you would get that the integral of a polynomial order $n$ equals $Kx$ plus a polynomial of order $n-1$, which is incorrect. $\endgroup$– Carlo BeenakkerCommented Mar 29, 2021 at 10:51
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$\begingroup$ I guess it might still be correct for $a\neq0$. $\endgroup$– RengarJGCommented Mar 29, 2021 at 11:03
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1 Answer
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This is a special case of a technique known as Hermite reduction (here is the original article from 1872). It can be derived by means of the following identity, $$\frac{d}{dx}\left(x^{n-1}\sqrt{ax^2+bx+c}\right)=\frac{c (n-1) x^{n-2}+b(n-1/2) x^{n-1}}{ \sqrt{a x^2+b x+c}}+\frac{a n x^n}{\sqrt{a x^2+b x+c}}.$$ So for $a\neq 0$ and $n\geq 1$ we have $$\int \frac{x^n}{\sqrt{ax^2+bx+c}}dx=\frac{1}{an}x^{n-1}\sqrt{ax^2+bx+c}+\int\frac{P_{n-1}}{\sqrt{a x^2+b x+c}}.$$ We can now repeat this reduction on the integral with the polynomial $P_{n-1}(x)$ in the numerator, until we reach $P_0$, producing the identity in the OP.
answered Mar 29, 2021 at 19:26