00:01
Problem we are solving the anti -derivative of one plus x squared over x dx.
00:12
For this problem, we're going to need to do some trig substitution, and our x is going to be tangent of theta.
00:27
That means dx is secant squared of theta, d theta.
00:34
And then let's just plug in our changed variables here.
00:42
Then we have integral of 1 plus tangent squared theta, all over tan theta times secant squared theta, d theta.
01:01
That we can simplify a little bit because the top there is just going to be, that is secant squared inside the square root.
01:18
And then the square root of secant squared is just secant over tanned theta times secant squared theta d theta.
01:46
So this here is, well, secant theta is 1 over cosine.
01:55
Tangent theta is sine over cosine.
01:58
So this can really be rewritten as the integral of 1 over sine theta times secant squared theta theta which is the same thing as co -secant.
02:26
So continuing on because we still can't integrate this directly.
02:33
We're going to turn that secant squared into something else.
02:38
We're going to make it into 1 plus tangent squared.
02:40
So then we're going to have co -secant theta times one plus tan squared theta d -theta.
02:51
If we distribute that co -sicant, then we'll get co -secant theta plus co -cicant times tangent squared is secant theta -tan theta.
03:15
So now we can split these up into two...