How can I integrate 7/(x^2-6x+18)?

Question

Steps Please!

Answer 1

Splitting the denominator to complete the square:
x² - 6x + 18 = x² - 6x + 9 + 9 =
(x - 3)² + 3²

Therefore,
∫7dx/(x² - 6x + 18) = ∫7dx/((x - 3)² + 3²) =
7∫(⅓² dx)/((x - 3)²/3² + 3²/3²) =
7/3 ∫(⅓ dx)/(((x - 3)/3)² + 1)

Why move the constants around like that? Because now if we let u = (x - 3)/3, so du = ⅓ dx, we can substitute and get:
7/3 ∫du/(u² + 1) =
7/3 arctan u + C =
7/3 arctan ((x - 3)/3) + C

Answer 2

You can use arctan to integrate this problem:

1) complete the square in the denominator
this will give you (x-3)^2 + 9 in the denominator

2) pull the seven to the front of the problem
3) integrate to get [1/3 arctan ((x-3)/3)]
4) put it all together: [7/3 arctan ((x-3)/3)]

Answer 3

type in wolfsfram integrator in a search engine. it works like a charm