# Thread: log rule for integration confusion

1. ## log rule for integration confusion

If I have a function:

f(x) = 1 / (x+1), and I want to find it's indefinite integral, I can apply the log rule and get:

F(x) = ln |x+1| + C

If, however, I multiply both the numerator and the denominator by 2, before integrating, I get:

f(x) = 1 / (x+1) = 2 / 2x + 2

F(x) = ln |2x+2| + C

Clearly, f(x) are equal in both cases, but I can't see how ln |2x+2| equals ln |x+1|

Any explanation? Thanks

2. ## Re: log rule for integration confusion

Originally Posted by gralla55
If I have a function:

f(x) = 1 / (x+1), and I want to find it's indefinite integral, I can apply the log rule and get:

F(x) = ln |x+1| + C

If, however, I multiply both the numerator and the denominator by 2, before integrating, I get:

f(x) = 1 / (x+1) = 2 / 2x + 2

F(x) = ln |2x+2| + C

Clearly, f(x) are equal in both cases, but I can't see how ln |2x+2| equals ln |x+1|

Any explanation? Thanks
Let $u = 2x+2$

therefore $du = 2 dx \leftrightarrow dx = \dfrac{du}{2}$ and $\int \dfrac{2}{2x+2} dx = \int \left(\dfrac{2}{u} \cdot \dfrac{du}{2}\right)$

We can cancel those two's to give $\int \dfrac{du}{u}$ and continue as normal from there.

3. ## Re: log rule for integration confusion

If you write: $\ln|2x+2|+C=\ln|2(x+1)|+C=\ln(2)+\ln|x+1|+C$
Because $\ln(2)$ is also an constant number you can say:
$\ln|x+1|+C'$
with $C'$ a new constant integration term.

4. ## Re: log rule for integration confusion

Originally Posted by gralla55
f(x) = 1 / (x+1), and I want to find it's indefinite integral, I can apply the log rule and get: F(x) = ln |x+1| + C

F(x) = ln |2x+2| + C
Clearly, f(x) are equal in both cases, but I can't see how ln |2x+2| equals ln |x+1|
Your mistake in using the same constant, i.e. they are different constants.

Recall that $\ln(|2x+2|)=\ln(|x+1|)+ln(2)$.
If $C$is the constant in the first then $C+\ln(2)$ is in the second.

5. ## Re: log rule for integration confusion

Thanks! Now I get it.