Evaluate $\int \frac {dx}{1+ \sqrt{x+1} + \sqrt {x+2} + \sqrt {x-1} }$

Question

How do we evaluate $$ \int \dfrac {dx}{1+ \sqrt{x+1} + \sqrt {x+2} + \sqrt {x-1} }\;\;? $$

Please help

Answer 1

Hint:

$\int\dfrac{dx}{1+\sqrt{x+1}+\sqrt{x+2}+\sqrt{x-1}}$

$=\int\dfrac{dx}{1+\sqrt{x+2}+\sqrt{x+1}+\sqrt{x-1}}$

$=\int\dfrac{1+\sqrt{x+2}-\sqrt{x+1}-\sqrt{x-1}}{(1+\sqrt{x+2}+\sqrt{x+1}+\sqrt{x-1})(1+\sqrt{x+2}-\sqrt{x+1}-\sqrt{x-1})}dx$

$=\int\dfrac{1+\sqrt{x+2}-\sqrt{x+1}-\sqrt{x-1}}{(1+\sqrt{x+2})^2-(\sqrt{x+1}+\sqrt{x-1})^2}dx$

$=\int\dfrac{1+\sqrt{x+2}-\sqrt{x+1}-\sqrt{x-1}}{1+2\sqrt{x+2}+x+2-(x+1+2\sqrt{x+1}\sqrt{x-1}+x-1)}dx$

$=\int\dfrac{1+\sqrt{x+2}-\sqrt{x+1}-\sqrt{x-1}}{3-x+2\sqrt{x+2}-2\sqrt{x^2-1}}dx$

$=\int\dfrac{(1+\sqrt{x+2}-\sqrt{x+1}-\sqrt{x-1})(3-x-2\sqrt{x+2}+2\sqrt{x^2-1})}{(3-x+2\sqrt{x+2}-2\sqrt{x^2-1})(3-x-2\sqrt{x+2}+2\sqrt{x^2-1})}dx$

$=\int\dfrac{(1+\sqrt{x+2}-\sqrt{x+1}-\sqrt{x-1})(3-x-2\sqrt{x+2}+2\sqrt{x^2-1})}{(3-x)^2-(2\sqrt{x+2}-2\sqrt{x^2-1})^2}dx$

$=\int\dfrac{(1+\sqrt{x+2}-\sqrt{x+1}-\sqrt{x-1})(3-x-2\sqrt{x+2}+2\sqrt{x^2-1})}{9-6x+x^2-4(x+2-2\sqrt{x+2}\sqrt{x^2-1}+x^2-1)}dx$

$=\int\dfrac{(1+\sqrt{x+2}-\sqrt{x+1}-\sqrt{x-1})(3-x-2\sqrt{x+2}+2\sqrt{x^2-1})}{5-10x-3x^2+8\sqrt{(x+2)(x^2-1)}}dx$

$=\int\dfrac{(1+\sqrt{x+2}-\sqrt{x+1}-\sqrt{x-1})(3-x-2\sqrt{x+2}+2\sqrt{x^2-1})(5-10x-3x^2-8\sqrt{(x+2)(x^2-1)})}{(5-10x-3x^2+8\sqrt{(x+2)(x^2-1)})(5-10x-3x^2-8\sqrt{(x+2)(x^2-1)})}dx$

$=\int\dfrac{(1+\sqrt{x+2}-\sqrt{x+1}-\sqrt{x-1})(3-x-2\sqrt{x+2}+2\sqrt{x^2-1})(5-10x-3x^2-8\sqrt{(x+2)(x^2-1)})}{(5-10x-3x^2)^2-64(x+2)(x^2-1)}dx$

$=\int\dfrac{(1+\sqrt{x+2}-\sqrt{x+1}-\sqrt{x-1})(3-x-2\sqrt{x+2}+2\sqrt{x^2-1})(5-10x-3x^2-8\sqrt{(x+2)(x^2-1)})}{9x^4-4x^3-58x^2-36x+153}dx$