This is one of my favorite integrals.
I posted a long discourse on it on SOSMath in 2003.
(1) Plato had the best method: .multiply top and bottom by
His answer: .
(2) CaptainBlack also had a good approach: Partial Fractions.
He had: .
(3) We can use Long Division:. .
(4) One day, while totally bored, I came up with this substitution . . .
Since , then is in a right triangle
. . with: . and
Hence: . . . . and: .
And we have: .
And, of course, you are expected to show that these answers are equivalent.
Well if you really want to be ugly I have another way of doing this.
And bring to standard form of ones of the know functions.
I leave the details to you.