# integral of a function in the real line.

$$\int_{a}^{b}{\left(\frac{1}{x^2} +1\right)^{\frac{1}{2}}dx}$$ for $0<a<b$.

I don't know how to compute this at all.

Can you give me a hint please?

The integral $$\int_{a}^{b}{\left(\frac{1}{x^2} +1\right)^{\frac{1}{2}}dx}$$ when $0<a<b$ can be written as $$\int_{a}^{b}{\frac{\sqrt{1+x^2}}{x}dx}=\int_a^bx^{-1}\sqrt{1+x^2}dx$$ Now set $1+x^2=t^2$.