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Step-by-Step Solution
Step 1: Rationalize the integrand
We need to evaluate the definite integral
$$
\int_{0}^{1} \frac{1}{\sqrt{3+x} + \sqrt{1+x}} \, dx.
$$
To simplify the integrand, multiply the numerator and denominator by
$ \sqrt{3+x} - \sqrt{1+x} $:
$$
\frac{1}{\sqrt{3+x} + \sqrt{1+x}}
\times
\frac{\sqrt{3+x} - \sqrt{1+x}}{\sqrt{3+x} - \sqrt{1+x}}
=
\frac{\sqrt{3+x} - \sqrt{1+x}}
{(\sqrt{3+x} + \sqrt{1+x})(\sqrt{3+x} - \sqrt{1+x})}.
$$
Observe that
$$
(\sqrt{3+x} + \sqrt{1+x})(\sqrt{3+x} - \sqrt{1+x}) = (3+x) - (1+x) = 2.
$$
Hence, the integrand becomes
$$
\frac{\sqrt{3+x} - \sqrt{1+x}}{2}.
$$
Step 2: Split the integral
Rewrite the integral accordingly:
$$
\int_{0}^{1} \frac{1}{\sqrt{3+x} + \sqrt{1+x}} \, dx
=
\int_{0}^{1}
\frac{\sqrt{3+x} - \sqrt{1+x}}{2}
\, dx
=
\frac{1}{2}
\int_{0}^{1}
\bigl(\sqrt{3+x} - \sqrt{1+x}\bigr)
\, dx.
$$
Step 3: Separate into two integrals
Distribute the integral:
$$
\frac{1}{2}
\int_{0}^{1} \sqrt{3+x}\,dx
\;-\;
\frac{1}{2}
\int_{0}^{1} \sqrt{1+x}\,dx.
$$
We will evaluate each integral separately, using the formula
$$
\int (k + x)^{\frac{1}{2}} \, dx
= \frac{2}{3} (k + x)^{\frac{3}{2}} + C.
$$
Step 4: Evaluate the first integral
Consider
$$
\int_{0}^{1} \sqrt{3+x} \, dx.
$$
Using
$$
\int \sqrt{3+x} \, dx
= \frac{2}{3} (3+x)^{\frac{3}{2}} + C,
$$
apply the limits from 0 to 1:
$$
\int_{0}^{1} \sqrt{3+x} \, dx
=
\left[
\frac{2}{3} (3+x)^{\frac{3}{2}}
\right]_{0}^{1}
=
\frac{2}{3}
\Bigl(
(3+1)^{\frac{3}{2}}
-
(3+0)^{\frac{3}{2}}
\Bigr).
$$
Note that $4^{\frac{3}{2}} = 8$ and $3^{\frac{3}{2}} = 3\sqrt{3}$, so
$$
\int_{0}^{1} \sqrt{3+x}\, dx
=
\frac{2}{3}
\bigl(
8
-
3\sqrt{3}
\bigr)
=
\frac{16}{3}
-
2\sqrt{3}.
$$
Step 5: Evaluate the second integral
Now consider
$$
\int_{0}^{1} \sqrt{1+x} \, dx.
$$
Using
$$
\int \sqrt{1+x} \, dx
=
\frac{2}{3} (1+x)^{\frac{3}{2}} + C,
$$
apply the limits from 0 to 1:
$$
\int_{0}^{1} \sqrt{1+x}\, dx
=
\left[
\frac{2}{3} (1+x)^{\frac{3}{2}}
\right]_{0}^{1}
=
\frac{2}{3}
\Bigl(
(1+1)^{\frac{3}{2}}
-
(1+0)^{\frac{3}{2}}
\Bigr).
$$
Since $2^{\frac{3}{2}} = 2\sqrt{2}$,
$$
\int_{0}^{1} \sqrt{1+x}\, dx
=
\frac{2}{3}
\bigl(
2\sqrt{2} - 1
\bigr)
=
\frac{4\sqrt{2}}{3}
-
\frac{2}{3}.
$$
Step 6: Combine the results
Substituting these back, we have
$$
\int_{0}^{1}
\frac{1}{\sqrt{3+x} + \sqrt{1+x}}
\, dx
=
\frac{1}{2}
\Bigg[
\Bigl( \frac{16}{3} - 2\sqrt{3} \Bigr)
-
\Bigl( \frac{4\sqrt{2}}{3} - \frac{2}{3} \Bigr)
\Bigg].
$$
Inside the brackets:
$$
\left(\frac{16}{3} - 2\sqrt{3}\right)
-
\left(\frac{4\sqrt{2}}{3} - \frac{2}{3}\right)
=
\frac{16}{3}
- 2\sqrt{3}
-
\frac{4\sqrt{2}}{3}
+
\frac{2}{3}
=
\frac{18}{3}
- 2\sqrt{3}
- \frac{4\sqrt{2}}{3}
=
6
- 2\sqrt{3}
- \frac{4\sqrt{2}}{3}.
$$
Multiplying by $ \tfrac{1}{2} $:
$$
\int_{0}^{1}
\frac{1}{\sqrt{3+x} + \sqrt{1+x}}
\, dx
=
\frac{1}{2}
\left(
6
- 2\sqrt{3}
- \frac{4\sqrt{2}}{3}
\right)
=
3
- \sqrt{3}
- \frac{2\sqrt{2}}{3}.
$$
Step 7: Identify the coefficients and compute the final value
Express this result as
$$
a + b\sqrt{2} + c\sqrt{3}.
$$
Here,
$$
a = 3, \quad b = -\frac{2}{3}, \quad c = -1.
$$
We want
$$
2a + 3b - 4c
=
2 \times 3
+ 3 \times \Bigl(-\frac{2}{3}\Bigr)
- 4 \times (-1).
$$
Simplify:
$$
2 \times 3 = 6,
\quad
3 \times \Bigl(-\frac{2}{3}\Bigr) = -2,
\quad
-4 \times (-1) = +4.
$$
Therefore,
$$
2a + 3b - 4c = 6 - 2 + 4 = 8.
$$
The required value is 8.
