Evaluate the integral in terms of inverse hyperbolic functions and natural logarithms

Question

Evaluate the integral in terms of a) inverse hyperbolic functions, and b) natural logarithms.

$$\int_0^{2\sqrt3} \frac{dx}{\sqrt{4+x^2}}$$

I solved a) but I can't figure out how to do b). I don't even know where to start. Any help would be appreciated.

Answer 1

Hints:

The basic formulae to know are

• $\DeclareMathOperator\ash{argsinh}\displaystyle\int\frac{\mathrm d\mkern1.5mu x}{\sqrt{x^2+1}}=\ash x,\;$ from which we deduce by substitution $$\int\frac{\mathrm d\mkern1.5mu x}{\sqrt{x^2+a^2}}=\frac1a\ash\Bigl(\frac xa\Bigr).$$
• $\ash x=\ln\bigl(x+\sqrt{x^2+1}\bigr)$.

For the last formula, $y=\ash x$ means $\sinh y =\dfrac{\mathrm e^{2y}-1} {2\mathrm e^y}=x$. So $u=\mathrm e^y$ is the positive root of the quadratic equation $$u^2-2xu-1=0, \enspace\text{i.e.}\quad u=x+\sqrt{x^2+1},$$ whence the formula.

The definite integral is thus $$\frac12\ash\Bigl(\frac x2\Bigr)\Bigg\lvert_0^{2\sqrt3}=\frac12\,\ash\bigl(\sqrt3\bigr)=\frac12\ln(\sqrt3+2).$$