Question:

<p>Let <span class="math-tex">\({I}=\int_\limits{0}^{1} \frac{1}{\sqrt{3+x}+\sqrt{1+x}} d x\)</span><br /> <span class="math-tex">\(=\int_{0}^{1} \frac{\sqrt{3+x}-\sqrt{1+x}}{(3+x)-(1+x)} d x\)</span> [on rationalising]<br /> <span class="math-tex">\(=\int_{0}^{1}\left[\frac{\sqrt{3+x}-\sqrt{1+x}}{2}\right] d x\)</span><br /> <span class="math-tex">\(=\left.\frac{1}{2}\left\{\frac{(x+3)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{(1+x)^{\frac{3}{2}}}{\frac{3}{2}}\right\}\right|_{0} ^{1}\)</span><br /> <span class="math-tex">\(=\frac{1}{3}\left\{\left[(1+3)^{\frac{3}{2}}-(1+1)^{\frac{3}{2}}\right]-\left[(0+3)^{\frac{3}{2}}-(1+0)^{\frac{3}{2}}\right]\right\}\)</span><br /> <span class="math-tex">\(=\frac{1}{3}\{8-2 \sqrt{2}-3 \sqrt{3}+1\}\)</span><br /> <span class="math-tex">\(=\frac{1}{3}\{9-2 \sqrt{2}-3 \sqrt{3}\}=3-\frac{2}{\sqrt{3}} \sqrt{2}-\sqrt{3}\)</span><br /> <span class="math-tex">\(=a+b \sqrt{2}+c \sqrt{3}\)</span><br /> <span class="math-tex">\(\Rightarrow a=\frac{3}{0}, b=-\frac{2}{3}, {C}=-1\)</span><br /> So, <span class="math-tex">\(2 a+3 b-4 c=-2+4=8\)</span></p>