Put,

$x-3=4t \hspace{0.3cm}=\gt\hspace{0.3cm} x=3+4t \hspace{0.3cm}=\gt\hspace{0.3cm} dx= 4dt $

When,

$x=3,t=0, x=7,t=1$

$\int_{0}^{1}(4t)^{1/4}\,t^{1/4}(4-4t)^{1/4}\,4 dt $

$=\int_{0}^{1}(4)^{1/4}(t)^{1/4} (4-4t)^{1/4}dt $

$=4^{1/4+1/4+1} \int_{0}^{1}(t)^{1/4} (1-t)^{1/4}dt $

$=4^{3/2} \beta (\frac{5}{4} , \frac{5}{4})=8 \frac{\sqrt{5/4} \sqrt{5/4}}{\sqrt{\frac{5}{4}+\frac{5}{4}}} $

$=8\frac{\frac{1}{4} \sqrt{\frac{1}{4}}\,\frac{1}{4}\,\sqrt{\frac{1}{4}}}{\sqrt{\frac{5}{2}}}$

$=\frac{2}{3}\frac{(\sqrt{1/4})^{2}}{\sqrt{\pi}}$