If $\mathrm{I}=\int \frac{\mathrm{dx}}{\sqrt{2\mathrm{x}+3}+\sqrt{\mathrm{x}+2}}$, then I equals

$\mathrm{I}=\int \frac{\sqrt{2\mathrm{x}+3}-\sqrt{\mathrm{x}-2}}{\mathrm{x}+1}\mathrm{dx}={\mathrm{I}}_1-{\mathrm{I}}_2$ where ${\mathrm{I}}_1=\int \frac{\sqrt{2\mathrm{x}+3}}{\mathrm{x}+1}\mathrm{dx}\text{ and }{\mathrm{I}}_2=\int \frac{\sqrt{\mathrm{x}+2}}{\mathrm{x}+1}\mathrm{dx}\text{ In }{\mathrm{I}}_1,$

Put $2\mathrm{x}+3={\mathrm{t}}^2,\mathrm{x}+2={\mathrm{z}}^2$
$$\begin{gathered} {\mathrm{I}}_1=\int \frac{\sqrt{2\mathrm{x}+3}}{\mathrm{x}+1}\mathrm{dx} \\ =\int \frac{t}{{\displaystyle \frac{t^2-1}{2}}}tdt=2\int \left(1+\frac{1}{t^2-1}\right)dt=2t+\frac{1}{2}\log \left|\frac{t-1}{t+1}\right|+c \\ =2\sqrt{2\mathrm{x}+3}+\frac{1}{2}\log \left|\frac{\sqrt{2\mathrm{x}+3}-1}{\sqrt{2\mathrm{x}+3}+1}\right|+c \end{gathered}$$
$$\begin{gathered} {\mathrm{I}}_2=\int \frac{\sqrt{\mathrm{x}+2}}{\mathrm{x}+1}\mathrm{dx} \\ put \mathrm{x}+2=z^2 \\ \Rightarrow dx=2zdz \\ {I}_2=\int \frac{2z^{2}dz}{z^2-1}=2\int \left(1+\frac{1}{z^2-1}\right)dz=2\left(z+\frac{1}{2}\log \left|\frac{z-1}{z+1}\right|\right)+c \end{gathered}$$