`int(sqrt(x^2-a^2))/x dx`

Options

A. `sqrt(x^(2)-a^(2))-acos^(-1)((a)/(x))+c`
B. `xsqrt(x^(2)-a^(2))-(1)/(a)tan^(-1)((x)/(a))+c`
C. `sqrt(x^(2)-a^(2))+asec^(-1)((x)/(a))+c`
D. `sqrt(x^(2)-a^(2))+(1)/(x)sec^(-1)(x)+c`

Answer

A

Explanation

Let `I=int(sqrt(x^(2)-a^(2)))/(x)dx`
Putting `x=asecthetaimpliestheta=sec^(-1)((x)/(a))`
`impliesdx=asec theta tan theta d theta`
`thereforeI=int(sqrt(asec^(2)theta-a^(2)))/(asectheta))` (a sec `theta tan theta d theta`)
`=inta tan theta* tan theta d theta = int a tan^(2) theta d theta`
`=a int (sec^(2)theta-1)d theta=a[tan theta- theta]+C`
`=a[sqrt(sec^(2)theta-1)-theta]+C`
`=a[sqrt((x^(2))/(a^(2))-1)-sec^(-1)((x)/(a))]+C`
`implies sqrt(x^(2)-a^(2))-acos^(-1)((a)/(x))+C`