Given equation is
$$
\sqrt{x+1}-\sqrt{x-1}=\sqrt{4 x-1}
$$
On squaring both sides, we get
$$
(x+1)+(x-1)-2 \sqrt{x^{2}-1}=4 x-1
$$
$\Rightarrow \quad 2 x-2 \sqrt{x^{2}-1}=4 x-1$
$\Rightarrow \quad-2 \sqrt{x^{2}-1}=2 x-1$
Again, squaring both sides, we get
$$
4\left(x^{2}-1\right)=4 x^{2}+1-4 x
$$
$\Rightarrow \quad-4=+1-4 x$
$\Rightarrow \quad 4 x=5$
$\Rightarrow$
$$
x=\frac{5}{4}
$$
But, when we put $x=\frac{5}{4}$ in Eq. (i), we get
$$
\sqrt{\frac{5}{4}+1}-\sqrt{\frac{5}{4}-1}=\sqrt{4 \times \frac{5}{4}-1}
$$
$\Rightarrow$
$\sqrt{\frac{9}{4}}-\sqrt{\frac{1}{4}}=\sqrt{5-1} \Rightarrow \frac{3}{2}-\frac{1}{2}=2$
$\Rightarrow 1=2,$ which is not true.
Hence, no value of $x$ satisfy the given equation.