How to find reduction formula for $I_n=\int\frac{(px+q)^n}{\sqrt{ax+b}}dx$

Question

I have to find the reduction formula for the following :

$$\int\frac{(px+q)^n}{\sqrt{ax+b}}dx$$

I took this integral from wikipedia. By using parts this is what I got.

${2a^{-1}(px+q)^{n}\sqrt{ax+b}-2pna^{-1}\int\{px+q}^{n-1}\sqrt{ax+b}dx$

The answer is in this link in reduction formula table: https://en.m.wikipedia.org/wiki/Integration_by_reduction_formulae

Any help is appreciated. Thanks in advance.

Answer 1

Hint: Perform integration-by-parts suggested below

$$I_n= \int\frac{(px+q)^n}{\sqrt{ax+b}}dx =\frac{1}{a(n+\frac12)} \int \left( \frac{px+q}{ax+b}\right)^n d\left( (ax+b)^{n+\frac12}\right) $$

Answer 2

We first perform integration by parts on $\sqrt{ax+b}$. \begin{aligned} I_{n} &=\int \frac{(p x+q)^{n}}{\sqrt{a x+b}} d x \\ &=\frac{2}{a} \int(p x+q)^{n} d(\sqrt{a x+b}) \\ &=\frac{2(p x+q)^{n}}{a \sqrt{a x+b}}-\frac{2 pn}{a} \int(p x+q)^{n-1} \sqrt{a x+b} d x \end{aligned} Rationalization gives back our integrals. $$ \begin{aligned} I_{n} &=\frac{2(p x+q)^{n}}{a \sqrt{a x+b}}-\frac{2 pn}{a} \int \frac{(p x+q)^{n-1}(a x+b)}{\sqrt{a x+b}} d x \\ &=\frac{2(p x+q)^{n}}{a \sqrt{a x+b}}-\frac{2 pn}{a} \int \frac{(p x+q)^{n-1}\left[\frac{a}{p}(p x+q)+b-\frac{a q}{p}\right]}{\sqrt{a x+b}} d x \\ &=\frac{2(p x+q)^{n}}{a \sqrt{a x+b}}-\frac{2 pn}{a}\left(\frac{a}{p} I_{n}+\frac{b p-a q}{p} I_{n-1}\right) \end{aligned} $$ Rearranging gives $$ \begin{aligned} \left(2n+1\right) I_{n} &=\frac{2(p x+q)^{n}}{a \sqrt{a x+b}}+\frac{2 n(a q-bp)}{a } I_{n-1} \\ I_{n} &=\frac{2 p(p x+q)^{n}}{(2 n+1)\sqrt{a x+b}}+\frac{2 n(aq-b p)}{2 n+1} I_{n-1}, \end{aligned} $$

which is the reduction formula of $I_n$.