If _0^1 1√3+x+√1+x ~d x=a+b √2+c √3, where a, b, c are rational numbers, then 2 a+3 ~b-4 c…

Question

If $\int\limits_0^1 \frac{1}{\sqrt{3+x}+\sqrt{1+x}} \mathrm{~d} x=\mathrm{a}+\mathrm{b} \sqrt{2}+\mathrm{c} \sqrt{3}$, where $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are rational numbers, then $2 \mathrm{a}+3 \mathrm{~b}-4 \mathrm{c}$ is equal to :

Accepted Answer

$$\begin{aligned}
& \int_\limits0^1 \frac{1}{\sqrt{3+x}+\sqrt{1+x}} d x=\int_\limits0^1 \frac{\sqrt{3+x}-\sqrt{1+x}}{(3+x)-(1+x)} d x \
& \frac{1}{2}\left[\int_\limits0^1 \sqrt{3+x} d x-\int_\limits0^1(\sqrt{1+x}) d x\right]
\end{aligned}$$

$$\begin{aligned}
& \frac{1}{2}\left[2 \frac{(3+x)^{\frac{3}{2}}}{3}-\frac{2(1+x)^{\frac{3}{2}}}{3}\right]_0^1 \
& \frac{1}{2}\left[\frac{2}{3}(8-3 \sqrt{3})-\frac{2}{3}\left(2^{\frac{3}{2}}-1\right)\right] \
& \frac{1}{3}[8-3 \sqrt{3}-2 \sqrt{2}+1] \
& =3-\sqrt{3}-\frac{2}{3} \sqrt{2}=a+b \sqrt{2}+c \sqrt{3} \
& a=3, b=-\frac{2}{3}, c=-1 \
& 2 a+3 b-4 c=6-2+4=8
\end{aligned}$$

Solutions · 1