Can we compute integral $ \int_a^b \frac {1}{\sqrt{1+x^2}} dx$ without $\operatorname{arcsinh}$?

Question

I'd like to compute

Can we compute $J = \int_a^b \frac {1}{\sqrt{1+x^2}} dx$, without $\operatorname{arcsinh}$?

Using Maple gives answers with $\operatorname{arcsinh}$ or $\operatorname{arctanh}$. I've also tried integration by parts, but with no success. How to compute this integral (possibly without using any trigonometric/hyperbolic function)?

Answer 1

Hints for $I$ :

(1) try to sub $x=2\tan u$

(2) $\int \csc xdx = -\ln |\csc x + \cot x| + C$

Answer 2

In fact $I$ was easier than I thought: After a substitution $v=\sqrt{t^2+4}$ and a partial fraction decomposition, one has:

$\int (\frac 14 \frac 1{v-2} - \frac 14 \frac 1{v+2}) dv,$ which gives an answer involving $\ln(v\pm2)$.

Answer 3

Utilize the Euler substitution $\sqrt{1+x^2}=t+x$ to get $dx =-\frac{t+x}tdt$ and \begin{align} \int \frac {1}{\sqrt{1+x^2}} dx=- \int \frac {1}{t+x} \frac{t+x}t dt=-\ln t +C \end{align} Thus $$\int_a^b \frac {1}{\sqrt{1+x^2}} dx=-\ln\left(\sqrt{1+x^2}-x\right)\bigg|_a^b=\ln\frac{\sqrt{1+a^2}-a}{\sqrt{1+b^2}-b} $$