# Thread: Substitution and Integration by Parts Problem, I keep getting it almost right!

1. ## Substitution and Integration by Parts Problem, I keep getting it almost right!

The question (and Wolfram's Answer) is: integral ln(2x+1 - Wolfram|Alpha) (Sorry, I still don't know how to use laTex)

however, I, Wolfram and my textbook each have different answers. So, my first though was that it was just different forms but then I plugged in x = 5 and get different answers with each equation. Maybe the difference is due to the constant not being used?

Here is the answer on the back of the book:
1/2 * (2x+1) * ln(2x+1) - x + C

My work is attached.

Any help would be greatly appreciated as always!

2. If you multiply Wolfram's answer out:

$\frac{1}{2} (2 x+1) (\ln(2 x+1)-1)+K$

$= \frac{1}{2}(2x+1)\ln (2x+1) - \frac{1}{2}(2x+1) + K$

$=\frac{1}{2}(2x+1)\ln (2x+1) - x \underbrace{-\ \frac{1}{2} + K}_{\text{Call this C}}$

$=\frac{1}{2}(2x+1)\ln (2x+1) - x + C$

Happier now ? In other words, yes the two answers simply differ by a constant.

3. Hello s3a
Originally Posted by s3a
The question (and Wolfram's Answer) is: integral ln(2x+1 - Wolfram|Alpha) (Sorry, I still don't know how to use laTex)

however, I, Wolfram and my textbook each have different answers. So, my first though was that it was just different forms but then I plugged in x = 5 and get different answers with each equation. Maybe the difference is due to the constant not being used?

Here is the answer on the back of the book:
1/2 * (2x+1) * ln(2x+1) - x + C

My work is attached.

Any help would be greatly appreciated as always!
I think the book and Wolfram are both right. You have lost a factor of $\tfrac12$ on the last line of your working, before the final $\ln(2x+1)$ term.

Any other apparent differences in the answers are down to (a) re-arranging the expression (b) the constant of integration.

4. Note that

$\frac{d}{dx} f(x) \ln (f(x)) = (1 + \ln(f(x)) \cdot f'(x))$.

Hence,

$\frac{d}{dx} (2x + 1) \ln (2x + 1) = (1 + \ln(2x + 1)) \cdot 2$

$\int (1 + \ln(2x + 1)) \cdot 2 dx = 2 \int (1 + \ln(2x + 1))dx = 2 \int dx + 2 \int \ln(2x + 1) dx$

$(2x + 1) \ln (2x + 1) = 2 \int dx + 2 \int \ln(2x + 1) dx$

$(2x + 1) \ln (2x + 1) - 2 \int dx = 2 \int \ln(2x + 1) dx$

$\frac{1}{2}\left((2x + 1) \ln (2x + 1) - 2 \int dx \right) = \int \ln(2x + 1) dx$