karush said:
$$\displaystyle I=\frac{1}{2}x\sqrt{4x^2+9}+\frac{9}{4}\arsinh\left(\frac{2}{3}x\right)+C$$
$$\sinh\left(x\right)=y=\frac{e^{x}-e^{-x}}{2}$$
$$2y=e^{x}-e^{-x}\ \ \implies\ \ e^{2x}-2ye^{x}-1=0 $$
?
How does $\displaystyle\frac{9}{4}\arsinh\left(\frac{2}{3}x\right) $ become $\displaystyle\frac{9\ln\left({\sqrt{4x^2+9}}\right)+2x}{4 }$
Well, in short, it doesn't...
$\displaystyle \begin{align*} y &= \frac{9}{4}\,\textrm{arsinh}\,{\left( \frac{2}{3}\,x \right) } \\ \frac{4}{9}\,y &= \textrm{arsinh}\,{ \left( \frac{2}{3}\,x \right) } \\ \sinh{ \left( \frac{4}{9}\,y \right) } &= \frac{2}{3}\,x \\ \frac{1}{2}\,\left( \mathrm{e}^{\frac{4}{9}\,y} - \mathrm{e}^{-\frac{4}{9}\,y} \right) &= \frac{2}{3}\,x \\ \mathrm{e}^{\frac{4}{9}\,y} - \mathrm{e}^{-\frac{4}{9}\,y} &= \frac{4}{3}\,x \\ \mathrm{e}^{\frac{4}{9}\,y} \,\left( \mathrm{e}^{\frac{4}{9}\,y} - \mathrm{e}^{-\frac{4}{9}\,y} \right) &= \frac{4}{3}\,x\,\mathrm{e}^{\frac{4}{9}\,y} \\ \left( \mathrm{e}^{\frac{4}{9}\,y} \right) ^2 - \frac{4}{3}\,x\,\mathrm{e}^{\frac{4}{9}\,y} -1 &= 0 \\ \left( \mathrm{e}^{\frac{4}{9}\,y} \right) ^2 - \frac{4}{3}\,x\,\mathrm{e}^{\frac{4}{9}\,y} + \left(- \frac{2}{3}\,x \right) ^2 - \left( -\frac{2}{3}\,x \right) ^2 - 1 &= 0 \\ \left( \mathrm{e}^{\frac{4}{9}\,y} - \frac{2}{3}\,x \right) ^2 - \frac{4}{9}\,x^2 - 1 &= 0 \\ \left( \mathrm{e}^{\frac{4}{9}\,y } - \frac{2}{3} \,x \right) ^2 &= \frac{4}{9}\,x^2 + 1 \\ \left( \mathrm{e}^{\frac{4}{9}\,y} - \frac{2}{3} \, x \right) ^2 &= \frac{4\,x^2 + 9}{9} \\ \mathrm{e}^{\frac{4}{9}\,y} - \frac{2}{3}\,x &= \pm \frac{\sqrt{4\,x^2 + 9}}{3} \\ \mathrm{e}^{\frac{4}{9}\,y} &= \frac{2\,x \pm \sqrt{4\,x^2 + 9}}{3} \\ \frac{4}{9} \, y &= \ln{ \left( \frac{2\,x \pm \sqrt{ 4\,x^2 + 9 }}{3} \right) } \\ y &= \frac{9\ln{ \left( \frac{2\,x \pm \sqrt{ 4 \, x^2 + 9 }}{3} \right) }}{4} \end{align*}$
and in order for this to exist, we require $\displaystyle \begin{align*} y = \frac{9\ln{\left( \frac{2\,x + \sqrt{4\,x^2 + 9}}{3} \right) }}{4} \end{align*}$.