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#
