Evaluate : (i) `intsqrt(x^(2)-16)dx`
(ii) `intsqrt(4x^(2)-5)dx`
(iii) `intsqrt(17x^(2)-11)dx`

Explanation

We know that `intsqrt(x^2-a^(2))dx=(x)/(2)sqrt(x^(2)-a^(2))-(a^(2))/(2)log|x+sqrt(x^(2)-a^(2))|+C`.
`therefore` (i) `sqrt(x^(2)-16)dx=intsqrt(x^(2)-4^(2))dx`
`=(x)/(2)*sqrt(x^(2)-4^(2))-(16)/(2)log|x+sqrt(x^(2)-16)|+C`
`=(x)/(2)*sqrt(x^(2)-16)-8log|x+sqrt(x^(2)-16)|+C`.
(ii) `intsqrt(4x^(2)-5)dx=2intsqrt(x^(2)-(5)/(4))dx=2intsqrt(x^(2)-((sqrt(5))/(2))^(2))dx`
`=2*[(x)/(2)sqrt(x^(2)-(5)/(4))-(5)/(8)-log|x+sqrt(x^(2)-(5)/(4))|]+C`
`=xsqrt(x^(2)-(5)/(4))-(5)/(4)log|xsqrt(x^(2)-(5)/(4))|+C`.
(iii) `intsqrt(17x^(2)-11)dx=sqrt(17)*intsqrt(x^(2)-(11)/(17))dx`
`=sqrt(17)*{(x)/(2)sqrt(x^(2)-(11)/(17))-(11)/(34)log|x+sqrt(x^(2)-(11)/(17))|}+C` `=(x)/(2)sqrt(17x^(2)-11-)(11sqrt(17))/(34)log|x+sqrt(x^(2)-(11)/(17))|+C`.