Find: ∫ (√ x)/((x+1)(x-1)) d x

$$\begin{aligned} & I=\int \frac{\sqrt{x}}{(x+1)(x-1)} d x \\ & \begin{aligned} & \sqrt{x}=t \text { gives } I=\int \frac{2 t^{2}}{\left(t^{2}+1\right)\left(t^{2}-1\right)} d t \\ & \text { Let } \frac{2 t^{2}}{\left(t^{2}+1\right)\left(t^{2}-1\right)}=\frac{2 z}{(z+1)(z-1)} ; \text { where } t^{2}=z \\ & \text { we have } \frac{2 z}{(z+1)(z-1)}=\frac{1}{z+1}+\frac{1}{z-1} \\ & \Rightarrow I=\int \frac{1}{t^{2}+1} d t+\int \frac{1}{t^{2}-1} d t \\ &=\tan ^{-1} t+\frac{1}{2} \log \left|\frac{t-1}{t+1}\right|+c \\ &=\tan ^{-1} \sqrt{x}+\frac{1}{2} \log \left|\frac{\sqrt{x}-1}{\sqrt{x}+1}\right|+c \end{aligned} \end{aligned} $$