well the method in general is just

$\ell = \displaystyle{\int_0^1}\sqrt{1+(f^\prime(x))^2}~dx $

here $f(x)=x^{3/2}-1$

$f^\prime(x)=\dfrac 3 2 x^{1/2}$

$(f^\prime(x))^2= \dfrac 9 4 x$

so we must find the integral

$\ell = \displaystyle{\int_0^1}\sqrt{1+\dfrac 9 4 x}~dx$

$u=\dfrac 9 4 x$

$du=\dfrac 9 4 ~dx$

$\ell = \displaystyle{\int_0^1}\sqrt{1+\dfrac 9 4 x}~dx\Rightarrow \dfrac 4 9 \displaystyle{\int_0^{9/4}}\sqrt{1+u}~du $

$\ell = \dfrac 4 9 \dfrac 2 3 (1+u)^{3/2}|^{9/4}_0=\dfrac 8 {27} \left(\left(1+\dfrac 9 4\right)^{3/2} - 1\right)\approx 1.43971$