1. ## Integration of quadratic denominaters error

Hello

im stuck on the following integral

$\displaystyle\int^9_5 \dfrac {1}{10x-9-x^2} \, dx$

ill go through how iv done it which seems correct but then when i get to inputting the limits it produces math error on my calculater so i must have done something wrong.

$\displaystyle\int^9_5 \dfrac {1}{-(x^2-10x) -9} \, dx$

$\displaystyle\int^9_5 \dfrac {1}{-(x^2-10x +25) +25 -9} \, dx$

$\displaystyle\int^9_5 \dfrac {1}{-(x-5) +16} \, dx$

$\displaystyle\int^9_5 \dfrac {1}{4^2 - (x-5)^2} \, dx$

so since

$\displaystyle\int \dfrac {dx}{a^2 - x^2} = \dfrac{1}{a}arctanh \dfrac {x}{a} +C$ we get $\dfrac{1}{4}arctanh \dfrac {x-5}{4} +C$

but when x = 9 acrtanh 1 which produces and error and when x = 5 acrtanh 0 which is 0

any help on this would be most appreciated

2. Originally Posted by jordanrs
Hello

im stuck on the following integral

$\displaystyle\int^9_5 \dfrac {1}{10x-9-x^2} \, dx$

ill go through how iv done it which seems correct but then when i get to inputting the limits it produces math error on my calculater so i must have done something wrong.

$\displaystyle\int^9_5 \dfrac {1}{-(x^2-10x) -9} \, dx$

$\displaystyle\int^9_5 \dfrac {1}{-(x^2-10x +25) +25 -9} \, dx$

$\displaystyle\int^9_5 \dfrac {1}{-(x-5) +16} \, dx$

$\displaystyle\int^9_5 \dfrac {1}{4^2 - (x-5)^2} \, dx$

so since

$\displaystyle\int \dfrac {dx}{a^2 - x^2} = \dfrac{1}{a}arctanh \dfrac {x}{a} +C$ we get $\dfrac{1}{4}arctanh \dfrac {x-5}{4} +C$

but when x = 9 acrtanh 1 which produces and error and when x = 5 acrtanh 0 which is 0

any help on this would be most appreciated
There is nothing wrong with one of the limits giving a result of 0.

For the one where arctanh is not defined, I suggest using:

$\displaystyle \lim_{n \to \infty} \dfrac{1}{4}\text{arctanh} \dfrac {x-5}{4} +C \bigg|_5^n$

3. Maybe this an improper integral, which diverge

4. yeah good point about it being undefined / going of to infinity didnt think about that. cheers for the help