How to integrate $I = \int_{-a/2}^{a/2}\frac{1}{\sqrt{x^2 + b}}dx$

Question

$$I = \int_{-a/2}^{a/2}\frac{1}{\sqrt{x^2 + b}}dx$$

I tried to integrate by parts but failed. I think I'm supposed to change variables using a hyperbolic sine, but I don't know this method.

Answer 1

Assuming $a,b>0$ we have: $$ \int_{-a/2}^{a/2}\frac{dx}{\sqrt{x^2+b}}=2\int_{0}^{\frac{a}{2\sqrt{b}}}\frac{dx}{\sqrt{x^2+1}}=2\operatorname{arcsinh}\frac{a}{2\sqrt{b}}=\log\frac{a+\sqrt{a^2+4b}}{-a+\sqrt{a^2+4b}}.$$

Answer 2

Since $$\sinh x=\frac{e^x-e^{-x}}2$$ and $$\cosh x=\frac{e^{x}+e^{-x}}2$$ we have that $$\cosh^2 x-\sinh^2 x=1$$

Assuming that $b>0$, write $$\sqrt{x^2+b}=\sqrt b\cdot \sqrt{\left(\frac x{\sqrt b}\right)^2+1}$$

and now, $$\frac x{\sqrt b}=\sinh t$$

Answer 3

If $b=0$, then the problem is trivial.

If $b>0$, then you should take a look at the function $\ln(x+\sqrt{1+x^2})$ (hyperbolic arcsin, aka areasinus). Its derivative is equal to $1/\sqrt{1+x^2}$ (found in any reference book on derivatives/primitives).

If $b<0$, then you should take a look at $\ln(\sqrt{x^2-1}+x)$ (see comment above).