If \(\int_0^1 \frac{1}{\sqrt{3+x}+\sqrt{1+x}} d x=a+b \sqrt{2}+c \sqrt{3}\), where \(a, b, c\) are rational numbers, the…

Let \(I=\int_0^1 \frac{1}{\sqrt{3+x}+\sqrt{1+x}}dx\).

\(I=\frac{1}{2}\int_0^1(\sqrt{3+x}-\sqrt{1+x})dx\).

Now \(I=\frac{1}{2}\left[\int_0^1\sqrt{3+x}\,dx-\int_0^1\sqrt{1+x}\,dx\right]\).

Using \(\int \sqrt{x+a}\,dx=\frac{2}{3}(x+a)^{3/2}\),

\(\int_0^1\sqrt{3+x}\,dx=\left[\frac{2}{3}(x+3)^{3/2}\right]_0^1=\frac{2}{3}(8-3\sqrt3)\).

Also \(\int_0^1\sqrt{1+x}\,dx=\left[\frac{2}{3}(x+1)^{3/2}\right]_0^1=\frac{2}{3}(2\sqrt2-1)\).

Thus \(I=\frac{1}{2}\left[\frac{2}{3}(8-3\sqrt3)-\frac{2}{3}(2\sqrt2-1)\right]\).

So \(I=\frac{1}{3}(8-3\sqrt3-2\sqrt2+1)\).

Therefore \(I=3-\frac{2}{3}\sqrt2-\sqrt3\).

Comparing with \(a+b\sqrt2+c\sqrt3\), we get \(a=3\), \(b=-\frac{2}{3}\), and \(c=-1\).

Now \(2a+3b-4c=2(3)+3\left(-\frac{2}{3}\right)-4(-1)\).

So \(2a+3b-4c=6-2+4=8\).