a$.$ Use trigonometric substitution to evaluate the integral $\backslash$int $(\backslash$sqrt$($x$^(2)-1))/($x$)$dx b$.$ In the table of integrals, we found that $\backslash$int $(\backslash$sqrt$($u$^(2)-$a$^(2)))/($u$)$du$=\backslash$sqrt$($u$^(2)-$a$^(2))-$acos$^(-1)(($a$)/(|$u$|))+$C and in another $\backslash$int $(\backslash$sqrt$($u$^(2)-$a$^(2)))/($u$)$du$=\backslash$sqrt$($u$^(2)-$a$^(2))-$asec$^(-1)((|$u$|)/($a$))+$C How would you reconcile these two answers? c$.$ Use either integral in part $($b$)$ to evaluate the integral $\backslash$int $\backslash$sqrt$($e$^(2$x$)-6)$dx