Question

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

Answer 1

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

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

$\int \frac{\sqrt{4+x^2}}{x^6}dx=-\int t^3\sqrt{4t^2+1}dt=-t^2\int t\sqrt{4t^2+1}dt+\int \frac{d\left(t^2\right)}{dt}(\int t\sqrt{4t^2+1}dt)dt$

$=-\frac{1}{12}{t}^2(4t^2+1{)}^{3/2}+\frac{1}{6}\int t(4t^2+1{)}^{3/2}dt$

$=-\frac{1}{12}{t}^2(4t^2+1{)}^{3/2}+\frac{1}{120}(4t^2+1{)}^{5/2}+C=\frac{(4+x^2{)}^{3/2}(x^2-6)}{120x^5}+C$

$a=4,b=-6⟹a+b=-2$