Question

$\int \frac{\sqrt{4+x^2}}{x^6}dx=\frac{{(a+x^2)}^{3/2}(x^2-b)}{120x^5}+C$ then $a+b$ equals to:

Answer 1

$\int \frac{\sqrt{4+x^2}}{x^6}dx$

Put $x=\frac{1}{t}⟹dx=-\frac{1}{t^2}dt$

$\int \frac{\sqrt{4+\frac{1}{t^2}}}{\frac{1}{t^6}}\times \frac{-dt}{t^2}=-\int t^3\sqrt{4t^2+1}dt$

Put $4t^2+1=l^2⟹tdt=\frac{ldl}{4}$

$\int -\frac{l^2-1}{4}\times \frac{l^{2}dl}{4}=-\frac{1}{16}\int (l^4-l^2)dl=-\frac{l^3\left(3l^2-5\right)}{240}=\frac{(4+x^2{)}^{\frac{3}{2}}(x^2-\frac{3}{2})}{120x^5}$

$⟹a=4,b=\frac{3}{2}⟹a+b=4+\frac{3}{2}=\frac{11}{2}$