Find $\int\frac{(ax^2-b)}{x\sqrt{c^2x^2-(ax^2+b)^2}}dx$

Question

How to find $$\int\frac{(ax^2-b)}{x\sqrt{c^2x^2-(ax^2+b)^2}}dx$$ ?

Usually in such cases I use the substitution $x=1/t$ but here it doesn't seem to work.Any hints or ideas?

Answer 1

Let $$I =\int\frac{(ax^2-b)}{x\sqrt{c^2x^2-(ax^2+b)^2}}dx = \int\frac{ax^2-b}{x\cdot x \sqrt{c^2-(ax+\frac{b}{x})^2}}dx$$

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

Now Put $\left(ax+\frac{b}{x}\right) = t\;,$ Then $\displaystyle \left(a-\frac{b}{x^2}\right)dx = dt$

So $$I = \int\frac{1}{\sqrt{c^2-t^2}}dt = \sin^{-1}\left(\frac{c}{t}\right)+\mathcal{C}$$

So $$I = \int\frac{(ax^2-b)}{x\sqrt{c^2x^2-(ax^2+b)^2}}dx=\sin^{-1}\left(\frac{cx}{ax^2+b}\right)+\mathcal{C}$$