If int_0^1 frac{1}{sqrt{3+x}+sqrt{1+x}} d x=a | vedclass

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(D) Rationalize the integrand: $\int_0^1 \frac{\sqrt{3+x}-\sqrt{1+x}}{(3+x)-(1+x)} d x = \frac{1}{2} \int_0^1 (\sqrt{3+x}-\sqrt{1+x}) d x$ Evaluate th

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$8$

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(D) Rationalize the integrand: $\int_0^1 \frac{\sqrt{3+x}-\sqrt{1+x}}{(3+x)-(1+x)} d x = \frac{1}{2} \int_0^1 (\sqrt{3+x}-\sqrt{1+x}) d x$ Evaluate the integral: $\frac{1}{2} \left[ \frac{2}{3}(3+x)^{3/2} - \frac{2}{3}(1+x)^{3/2} \right]_0^1$ $= \frac{1}{3} \left[ (3+x)^{3/2} - (1+x)^{3/2} \right]_0^1$ $= \frac{1}{3} \left[ (4^{3/2} - 2^{3/2}) - (3^{3/2} - 1^{3/2}) \right]$ $= \frac{1}{3} \left[ (8 - 2\sqrt{2}) - (3\sqrt{3} - 1) \right] = \frac{1}{3} [9 - 2\sqrt{2} - 3\sqrt{3}] = 3 - \frac{2}{3}\sqrt{2} - \sqrt{3}$ Comparing with $a+b\sqrt{2}+c\sqrt{3}$,we get $a=3$,$b=-\frac{2}{3}$,$c=-1$ Calculate $2a+3b-4c = 2(3) + 3(-\frac{2}{3}) - 4(-1) = 6 - 2 + 4 = 8$

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,then $2 a+3 b-4 c$ is equal to :

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