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I'm trying to find a reduction formula for:

$${I}_n=\int \frac{x^{n}}{\sqrt{ax+b}}dx$$

This is where I have gotten to so far:

$${I}_n= \frac{2x^{n}\sqrt{ax+b}}{a}-\frac{2n}{a} \int \frac{x^{n-1}(ax+b)}{\sqrt{ax+b}}dx$$

I'd appreciate a little nudge in the right direction :)

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You're almost there. Simply note that for $I_n=\int \frac{x^n}{\sqrt{ax+b}}\,dx$, we have

$$\begin{align} I_n&=\frac{2x^n\sqrt{ax+b}}{a}-\frac{2n}{a}\int x^{n-1}\sqrt{ax+b}\,dx\\\\ &=\frac{2x^n\sqrt{ax+b}}{a}-\frac{2n}{a}\int x^{n-1}\frac{ax+b}{\sqrt{ax+b}}\,dx\\\\ &=\frac{2x^n\sqrt{ax+b}}{a}-\frac{2n}{a}\left(a\int \frac{x^n}{\sqrt{ax+b}}\,dx+b\int \frac{x^{n-1}}{\sqrt{ax+b}}\right)\\\\ &=\frac{2x^n\sqrt{ax+b}}{a}-2nI_n-\frac{2nb}{a}I_{n-1}\\\\ I_n&=\frac{2x^n\sqrt{ax+b}}{a(1+2n)}-\frac{2nb}{a(2n+1)}I_{n-1} \end{align}$$

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  • $\begingroup$ Simple and nice ! Cheers, my friend. $\endgroup$ – Claude Leibovici Jan 26 '17 at 7:18
  • $\begingroup$ @ClaudeLeibovici Thank you my friend! $\endgroup$ – Mark Viola Jan 26 '17 at 14:42
  • $\begingroup$ @FelixMarin Thank you Felix! Much appreciate. $\endgroup$ – Mark Viola Jan 28 '17 at 21:01

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