With this general case, notice how #sqrt(x^2 - a^2) prop sqrt(sec^2theta - 1)#. So, we can use the following substitution:
#x = asectheta#
#dx = asecthetatanthetad theta#
#sqrt(x^2 - a^2) = sqrt(a^2sec^2theta - a^2) = atantheta#
Thus, we have:
#= int 1/(cancel(atantheta))*cancel(a)secthetacancel(tantheta)d theta#
#= int secthetad theta#
Then just a little trick:
#= int sectheta((sectheta + tantheta)/(sectheta + tantheta))d theta#
#= int (sec^2theta + secthetatantheta)/(sectheta + tantheta)d theta#
Now, let:
#u = sectheta + tantheta#
#du = secthetatantheta + sec^2thetad theta#
Therefore:
#= int1/udu#
#= ln|u|#
#= ln|sectheta + tantheta|#
Recall that:
#sectheta = x/a#
#tantheta = sqrt(x^2 - a^2)/a#
Thus we have:
#= color(green)(ln|x/a + sqrt(x^2 - a^2)/a| + C)#
...which is perfectly acceptable. But, you could simplify this more and be a little sneaky.
#= ln|(1/a)[x + sqrt(x^2 - a^2)]| + C#
#= ln|x + sqrt(x^2 - a^2)| + ln|1/a| + C#
but since #a# is a constant... it gets embedded into #C#.
#= color(blue)(ln|x + sqrt(x^2 - a^2)| + C)#
So if you see Wolfram Alpha give you this answer, that's why.
