**You can see the final answer** **here**.

This is going to end up using the following major pieces of knowledge:

- Completing the Square
- Trigonometric Substitution
- The trick to #intsecthetad theta# (involves u-substitution)
- Properties of Logarithms

#int 1/sqrt(t^2 + t - 2)dt#

#t^2 + t - 2 = (t - 1)(t + 2)#

Although you CAN factor this, I'm going to **complete the square** instead. I want to get rid of the square root.

#t^2 + t - 2 = t^2 + t + 1/4 - 1/4 - 2 = (t + 1/2)^2 - 9/4#

#=> int 1/sqrt((t + 1/2)^2 - 9/4)dt#

Surprisingly, now we have the form:

#sqrt(x^2 - a^2)#

Now things are looking up for **Trigonometric Substitution**. I was taught to use #x = a"trig"theta#, so with #x = t + 1/2# and #a = 3/2#, let:

#t + 1/2 = 3/2sectheta#

#(t + 1/2)^2 = 9/4sec^2theta#

#dt = 3/2secthetatanthetad theta#

#sqrt((t+1/2)^2 - 9/4) = sqrt(9/4sec^2theta - 9/4) = 3/2sqrt(sec^2theta - 1) = 3/2sqrt(tan^2theta) = 3/2tantheta#

Now, we have:

#=> int 1/(cancel(3/2)cancel(tantheta))cancel(3/2)secthetacancel(tantheta)d theta#

#= int secthetad theta#

Not so bad, I guess... but you have to know this integral by now. Not because I expect that you should know more, but **you'll see it eventually if you haven't already** (it's pretty popular...). At least, I've been taught to do this integral. If you haven't learned it yet, here is how to do #intsecthetad theta#:

With malice aforethought... multiply by #(sectheta+tantheta)/(sectheta+tantheta)#.

#= 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#

#=> int 1/udu#

WOAH! Much easier now!

#= ln|u|#

#= ln|sectheta + tantheta| + C#

Going back to our substitutions:

#sectheta = 2/3(t+1/2)#

#tantheta = 2/3sqrt((t+1/2)^2 - 9/4) = 2/3sqrt(t^2 + t - 2)#

So the final result is:

#= ln|2/3(t+1/2) + 2/3sqrt(t^2 + t - 2)| + C#

#= ln|1/3(2(t+1/2) + 2sqrt(t^2 + t - 2))| + C#

#= ln|1/3(2t + 1 + 2sqrt(t^2 + t - 2))| + C#

Now, with the **properties of logarithms**, separate #ln# out.

#= ln|2sqrt(t^2 + t - 2) + 2t + 1| + ln(1/3) + C#

but #ln(1/3) = -ln 3#, both of which are just constants. So now, we get a new constant #D# which we can embed into #C# and call it #C# again... oh well. It's all still a constant, and the point of it is to say that there are infinite vertical shifts of the antiderivative that can be differentiated to give #1/sqrt(t^2 + t - 2)#. In other words, the value of the constant is not relevant. Finally, we get:

#= color(blue)(ln|2sqrt(t^2 + t - 2) + 2t + 1| + C)#