# Math Help - Integrals leading to a Log function

1. ## Integrals leading to a Log function

Hi all,
I am working on a problem at the moment which i have the answer for but i don't understand one step of the solution.

Evaluate "integration sign" (2x-6)dx/[(x^2)-6x+10]

The answer is given as Ln2.

I can see how the answer comes about but a step of it is confusing me.

Once i substitue and solve i get Ln[1x^2)-6(x)+10]/??? as per the rule for integrating 1/ax+b=ln(ax+b)/a

In otherwords 1/ax+b=(1[ln(ax+b)]/a) +c

However which value do i pic for a? It seems there are 2 coefficients of x in this case, 1 for the x squared, and -6 for the x.

It looks like the a chosen in the solution is the coefficient of the X^2 which is 1.

Why is the coefficient of the -6x not used for the a ?

It seems a pretty fundemental issue that i need solved.

When given a quadratic expression & when integrating is the a coefficient for the solution always the coefficient associated with the x^2?

Apologies if this is a ridiculous question
John

2. ## Re: Integrals leading to a Log function

You're right that:
$\int \frac{2x-6}{x^2-6x+10}dx= \ln|x^2-6x+10|+C$
I think it's strange there're no integration limits given.

3. ## Re: Integrals leading to a Log function

I think we have to take t = [(x^2)-6x+10]. Then dt = 2x-6. If you substitute this, your problem will become integration of dt/t.

4. ## Re: Integrals leading to a Log function

sorry guys the limits are b=4 a=3

5. ## Re: Integrals leading to a Log function

That changes the cases, that means we have:
$[\ln|x^2-6x+10|]_{3}^{4}=\ln|4^2-24+10|-\ln|3^2-18+10|=\ln|2|-\ln|1|=\ln|2|-0=\ln(2)$

6. ## Re: Integrals leading to a Log function

i should have just solved the expression -tut!
thanks all
John

7. ## Re: Integrals leading to a Log function

But if you are going to make the substitution $t= x^2- 6x+ 10$, $dx= (2x- 6)dx$, you might as well change the limits of integration also: when x= 3, $t= 9- 18+ 10= 1$ and when x= 4 $t= 16- 24+ 10= 2$ so
$\int_3^4 \frac{2x- 6}{x^2- 6x+ 10}dx= \int_1^2\frac{dt}{t}= \left[ln(t)\right]_1^2= ln(2)- ln(1)= ln(2)$.