Prove a reduction formula for: $\int \frac{x^n dx}{\sqrt{ax^2 + bx + c}}$

Question

For $n > 1\in \Bbb N$, prove that: $$ J_n = \int \frac{x^n dx}{\sqrt{ax^2 + bx + c}} = \\ {1\over na}\left(x^{n-1}\sqrt{ax^2 + bx + c} - {b\over 2}(2n-1)J_{n-1} - c(n-1)J_{n-2}\right) $$

I've been working on this for a while without any success. I've first tried to use the fact that: $$ \int {P_n(x)dx\over \sqrt{ax^2 + bx + c}} = Q_{n-1}(x)\sqrt{ax^2 + bx + c} + \int {\lambda dx \over \sqrt{ax^2 + bx + c}} $$ where $Q_{n-1}(x)$ is a polymonial of $n-1$ degree at max and coeeficients for $Q_{n-1}(x)$ and $\lambda$ are yet to be determined. Using polynomials from the problem statement we get: $$ \int \frac{x^n dx}{\sqrt{ax^2 + bx + c}} = Ax^{n-1}\sqrt{ax^2 + bx + c} + \int{\lambda dx \over \sqrt{ax^2 + bx + c}} $$

Differente both parts of the equality, after which we need to find the coefficients: $$ \frac{x^n dx}{\sqrt{ax^2 + bx + c}} = {d(Ax^{n-1}\sqrt{ax^2 + bx + c})\over dx} + {\lambda \over \sqrt{ax^2 + bx + c}} = \\ A(n-1)x^{n-2}\sqrt{ax^2 + bx + c} + Ax^{n-1}\frac{2ax + b}{2\sqrt{ax^2 + bx + c}} + {\lambda dx \over \sqrt{ax^2 + bx + c}}$$

Multiplying both sides by $\sqrt{ax^2 + bx + c}$ and skipping some algebraic transformation I was indeed able to get that: $$ A = {1\over na} $$

However, the term $\lambda$ appears to be equal to $0$. Which yields: $$ J_n = {x^{n-1}\over na}\sqrt{ax^2 + bx + c} $$ And this approach doesn't seem to lead anywhere. I've then tried a different technique. Let $b = 2b_0$, then: $$ aJ_{n+2} = \int \frac{ax^{n+2}dx}{\sqrt{ax^2 + 2b_0x+c}}\\ 2b_0J_{n+1} = \int \frac{2b_0x^{n+1}dx}{\sqrt{ax^2 + 2b_0x+c}}\\ cJ_n = \int \frac{cx^{n}dx}{\sqrt{ax^2 + 2b_0x+c}} $$

Summing left and right parts we get: $$ aJ_{n+2} + 2b_0J{n+1} + cJ_n = \\ \int \frac{(ax^2 + 2b_0x + c)x^n dx}{\sqrt{ax^2 + 2b_0x + c}}= \\ \int {x^n\sqrt{ax^2 + 2b_0x + c}} dx $$

Integration by parts yields: $$ u = \sqrt{ax^2 + 2b_0x + c}\\ du = {2ax + 2b_0\over 2\sqrt{ax^2 + 2b_0x + c}}dx\\ dv = x^n\\ v = {x^{n+1}\over n+1}\\ \int {x^n\sqrt{ax^2 + 2b_0x + c}} dx = uv - \int vdu = \\ = {x^{n+1}\over n+1} \sqrt{ax^2 + 2b0x + c} - {J_{n+2}\over n+2} - {J_{n+1}\over 2(n+1)} $$

Which seems to be "the other way round". The question is what technique do I use to prove what's stated in the question section? Hopefully, I didn't make typos in the body. Thank you!

Answer 1

Hint:

$$\dfrac{d(x^m\sqrt{ax^2+bx+c})}{dx}$$

$$=mx^{m-1}\sqrt{ax^2+bx+c}+\dfrac{x^m(2ax+b)}{2\sqrt{ax^2+bx+c}}$$

Now let $x^3(2ax+b)=(2x^2+dx+e)(ax^2+bx+c)-ce$

Find $d,e$ by comparing the coefficients of $x,x^2,x^3$

Integrate both sides of the first relationship wrt $x$

Answer 2

Note that $$\left(x^{n-1} \sqrt{ax^2 + bx + c}\right)’ = \frac {nax^n+(n-\frac12)b x^{n-1}+(n-1)c x^{n-2}}{\sqrt{ax^2 + bx + c}} $$ and integrate both sides to obtain $$ J_n= \frac{x^{n-1}}{na}\sqrt{ax^2 + bx + c}-\frac{(2n-1)b}{2na} J_{n-1}-\frac{(n-1)c}{na} J_{n-2}$$

Answer 3

Late answer but there is an elegant way to derive the reduction formula as follows:

We have $$ aJ_n + \frac{b}{2}J_{n-1} = \int \frac{x^{n-1}(2ax+b)dx}{2\sqrt{ax^2+bx+c}} = \int x^{n-1} d\left(\sqrt{ax^2+bx+c}\right) $$

We then proceed by integrating by parts $$ \int x^{n-1} d\left(\sqrt{ax^2+bx+c}\right) = x^{n-1}\sqrt{ax^2+bx+c} - (n-1)\int x^{n-2}\sqrt{ax^2+bx+c} dx $$

To calculate $\int x^{n-2}\sqrt{ax^2+bx+c}$, note that

$$ aJ_n + bJ_{n-1} + cJ_{n-2} = \int \frac{x^{n-2}(ax^2+bx+c)}{\sqrt{ax^2+bx+c}} = \int x^{n-2}\sqrt{ax^2+bx+c} $$

Substituting back $$ aJ_n + \frac{b}{2}J_{n-1} = x^{n-1}\sqrt{ax^2+bx+c} - (n-1)\left(aJ_n + bJ_{n-1} + cJ_{n-2}\right) $$

By grouping terms, we easily get the reduction formula. QED.