Reduction formula for $\int \frac{dx}{x^n \sqrt{ax+b}}$

Question

I want a reduction formula for

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

in terms of $I_{n-1}$.

I have tried various substitutions but I just can't seem to find the right one. Any help or hints will be appreciated

Answer 1

Let $I_k$ be as in the OP. Integrate by parts, letting $u=x^{-k}$ and $dv=\frac{dx}{\sqrt{ax+b}}$. Then $du=-kx^{-k-1}$ and we can take $v=\frac{2}{a}\sqrt{ax+b}$. We get $$I_k =uv+\frac{2k}{a}\int \frac{\sqrt{ax+b}}{x^{k+1}}\,dx.$$

Now note that $$\int \frac{\sqrt{ax+b}}{x^{k+1}}\,dx=\int \frac{ax+b}{x^{k+1}\sqrt{ax+b}}\,dx=aI_k+bI_{k+1}.$$

Routine algebraic manipulation now gives us $I_{k+1}$ in terms of $I_k$. The above argument breaks down if $a=0$ or $b=0$, but then we don't need a reduction formula.

Answer 2

Here's a hint:

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

The second integral is already in terms of $I_{n-1}$. Use integration by parts on the first integral.