# Integration using natural logs, halp! (beginning of my calc 2 class)

1. Jan 20, 2013

### Vagabond7

1. The problem statement, all variables and given/known data
Ok, so I have the problem, and I have the answer, but I don't know how they arrived at the answer. I'll show you the problem, and what I tried and maybe somebody can point out where I am going wrong, or some hints to put me on the right path.

(2x-1)/(x+1) being integrated on ∫ 0 to 6

this lesson is supposed to be demonstrating how to integrate using natural log.

The answer I was given was

12-3 ln 7

3. The attempt at a solution

I tried a variety of things, none of which led me to the given solution. I tried breaking it up into (2x)/(x+1)-(1)/(x+1) and then integrating each piece (which I'm not sure is a "legal" move) getting x^2 ln (x+1)-x ln (x+1), or trying it as x^2-x ln (x+1) neither of which gets me to the final answer.

Now there was a hint to use "U substitution" but I had in the past been told that I can only do "u substitution" if the derivative is found elsewhere in the function or the derivative is off by a constant factor. the derivative of neither expression is the other off by a constant factor (they would both be off by a variable x), so I'm not sure how to do a "u substitution" in that case. Halp?

2. Jan 20, 2013

### Dick

Put u=(x+1). Then du=dx and x=(u-1). Substitute that into 2x-1 to get the numerator in terms of u.

3. Jan 20, 2013

### lurflurf

$$\int_0^6 \! \frac{2x-1}{x+1}\, \mathrm{dx}=\int_0^6 \! \frac{2(x+1)-3}{x+1}\, \mathrm{dx}=\int_0^6 \! \left(2-3\frac{1}{x+1} \right)\, \mathrm{dx}$$

4. Jan 24, 2013

### Vagabond7

Oh, I forgot to thank you guys. That totally cleared up that problem and helped alot with methods for a couple of other ones. Thanks gentlemen!