how to integrate $\int\frac{1}{x^2-12x+35}dx$?

Question

How to integrate following

$$\int\frac{1}{x^2-12x+35}dx?$$ What I did is here:

$$\int\frac{dx}{x^2-12x+35}=\int\frac{dx}{(x-6)^2-1}$$

substitute $x-6=t$, $dx=dt$ $$=\int\frac{dt}{t^2-1}$$ partial fraction decomposition, $$=\int{1\over 2}\left(\frac{1}{t-1}-\frac{1}{t+1}\right)dt$$ $$=\frac12(\ln|1-t|-\ln|1+t|)+c$$ $$=\frac12\left(\ln\left|\frac{1-t}{1+t}\right|\right)+c$$ substitute back to $t$ $$=\frac12\ln\left|\frac{7-x}{x-5}\right|+c$$

I am not sure if my answer correct. Can I integrate this without substitution? Thank you

Answer 1

Can you skip the substitution? Certainly!

$\frac {1}{x^2 - 12 x + 35} = \frac {1}{(x-7)(x-5)} = \frac {A}{x-7} + \frac {B}{x-5}$

$A+B = 0\\ -5A - 7B = 1\\ A = \frac {1}{2}, B = -\frac {1}{2}$

$12\int \frac {1}{x-7} - \frac {1}{x-5} \ dx\\ \frac 12 (\ln |x-7| - \ln |x-5|)\\ \frac 12 \ln \left|\frac {x-7}{x-5}\right|$

Answer 2

You can do that without substitution. Use partial fractions by factorizing denominator:$x^2-12x+35=(x-5)(x-7)$ $$\int \frac{dx}{x^2-12x+35}=\int \frac{dx}{(x-7)(x-5)}$$ $$=\int\frac12\left( \frac{1}{x-7}-\frac{1}{x-5}\right)dx$$ $$=\frac12(\ln\left| x-7\right|-\ln\left| x-5\right|)$$ $$=\frac12\ln\left| \frac{x-7}{x-5}\right|+C$$