00:01
For this problem, we are to evaluate the given integral for several values of k.
00:06
So let's start with the first case.
00:08
That is when k is equal to 0.
00:11
If k is zero, then we have 1 over x squared plus k dx.
00:17
It's going to be the integral of 1 over x squared dx.
00:22
And this is equal to the integral of x raised negative 2 dx.
00:27
And integrating that, we get x.
00:30
Raised to negative 2 plus 1.
00:32
It's a negative 1 over negative 2 plus 1, which is also negative 1, and then plus c, or that's the same as negative 1 over x plus c.
00:42
What if our k is greater than 0? if k is greater than 0, we have the integral of 1 over x squared plus k, dx.
00:53
And if i factor out the k in the denominator, i'll then have integral of 1 over x squared plus k, dx.
00:57
And if i factor out the k in the denominator, i'll then have integral of 1 over k, this times x squared over k plus 1 and then dx.
01:10
That's going to be equal to the integral of 1 over k times x over the square of k squared plus 1 and then d x.
01:22
And then in here we will do substitution.
01:25
You want to let u equal to x over the square of k.
01:29
That means du is equal to 1 over the square of k dx or square root of k times du is equal to dx so what happens now is that we have 1 over k times the integral of 1 over we have u squared plus 1 times the square of 1 times the square of k d u.
01:55
Multiplying now 1 over k and the square of k we should get 1 over the square of k integral of 1 over u squared plus 1 d u which is the same as 1 over the square of k tangent inverse of u plus c but since u is x over the square of k this is equal to 1 over the square of k tangent inverse of x or the square of k.
02:25
Over the squared of k and then plus c...