Question

using integration by substitution,what is the integral of x-3/x^2-6x+5

Answer 1

Let u = denominator

Answer 2

factorize the base so u would could cut the numberator and reduce it to a single denominator method

Answer 3

**numerator**

Answer 4

What I was saying is that you have something like:\[\int\limits_{a}^{b}\frac{x-3}{x^2-6x+5}dx\]Let\[u = x^{2} - 6x + 5\]and\[du = (2x - 6)dx\]You note that the numerator is 1/2 of what it should be. If it was 2(x-3), everything would be great.
Understood so far?

Answer 5

yes i do.

Answer 6

now its really easy to do @good one @bmp

Answer 7

But you can't just multiply it by 2 as that would change the integral. What is the neutral factor in multiplication? 1. So, we should only multiply it by 1.
\[\frac{2}{2} \int\limits_{a}^{b}\frac{x-3}{x^2-6x+5}dx = \frac{1}{2}\int_{a}^{b}\frac{2(x-3)}{x^{2}-6x+5}dx\]

Answer 8

Now, you are free to use the above substitution that you left you with:\[\frac{1}{2}\int\limits_{v}^{t}\frac{du}{u} = \frac{1}{2}\ln{|u|} + C\]