Answer
int 1/(x^2 - 2x + 17)"dx" = 1/4tan^-1((x - 1)/4) + C
Method
The idea behind this one is: you want to make that base( denominator: x^2 - 2x + 17 ) look something like 1/(x^2 + 1)
because the integral(primitive) of that is tan^-1x !
So this means that we can use the substitution of tan
You'll surely get that intuition as we move along!
Solution
x^2 - 2x + 17 = (x - 1)^2 - 1 + 17 By completing the square
= (x - 1)^2 + 16
= 16[((x - 1)^2)/16 + 1] = 16[((x - 1)/4)^2 + 1]
=> 1/(x^2 - 2x + 17) = 1/(16[((x - 1)/4)^2 + 1])
Therefore,
int 1/(x^2 - 2x + 17)"dx" = 1/16int 1/(((x - 1)/4)^2 + 1)"dx"
<Now that looks more like int1/(x^2 + 1)>
So we let (x - 1)/4 = tantheta
=> (dx)/4 = sec^2theta d theta => dx = 4sec^2theta d theta
=> 1/16int 1/(tan^2theta + 1)*4sec^2theta d theta
=> 4/16intcancel(sec^2theta)/cancel(sec^2theta) d theta
=> 1/4intd theta = 1/4theta + C
Remember we let, (x - 1)/4 = tantheta => theta = tan^-1((x - 1)/4)
=> 1/4tan^-1((x - 1)/4) + C
