Im solving an the integral of $\displaystyle x^2/sqrt(x^2-9)$ using trig substitution. I have it down to $\displaystyle 9*INT sec(x)^3 = secxtanx - INT sec(x)^3 + ln(secx + tanx)$ so I add the one integral over to the other side and get $\displaystyle 10*INT sec(x)^3 = secxtanx + ln(secx + tanx)$ and then I get $\displaystyle INT sec(x)^3 = 1/10[secxtanx + ln(secx+tanx)]$ but my solutions manual has $\displaystyle INT sec(x)^3 = 9/2[secxtanx + ln(secx+tanx)]$ I'm just confused as to why I can't add the other integral over to get 10. It seems with other integration by parts problems you can but for some reason with the sec(x)^3 they don't add it over?