Evaluate $\int \frac{a}{x(x^2-a^2)^{1/2}} dx$ I was trying to integrate $\int \frac{a}{x(x^2-a^2)^{1/2}} dx$ and by applying on what i saw on the formula of inverse trigonometric functions, there is formula like $\frac{1}{a}Arcsec\frac{u}{a}$ = $\int \frac{du}{u\sqrt(u^2-a^) }$ so my answer is $\frac{1}{a}Arcsec\frac{x}{a}$ but my friend said it should be something like $arcsin?$

$$I=\int\dfrac1{x\sqrt{x^2-a^2}}dx=\int\dfrac x{x^2\sqrt{x^2-a^2}}dx$$
Let $\sqrt{x^2-a^2}=u\implies\dfrac x{\sqrt{x^2-a^2}}dx=du$ and $x^2=u^2+a^2$
$$I=\int\dfrac{du}{u^2+a^2}=\dfrac1a\arctan\dfrac ua+K$$
Now we can prove $$\arctan y=\arcsin\dfrac y{\sqrt{1+y^2}}\text{ OR } \arcsin z=\arctan\dfrac z{\sqrt{1-z^2}}$$
and use $\arcsin a+\arccos a=\dfrac\pi2$ for $|a|\le1$
and $\arccos p=\text{arcsec}\dfrac1p$ for $|p|\le1$
