Integration Question

[Jason76]

\(\displaystyle \frac{x}{(4 - x^{2})}dx\)

\(\displaystyle du = -2x\)

which is later flipped as \(\displaystyle \frac{1}{2}\) (for a short time as negative, but then postive in the final answer)

Trying to get into this form:

\(\displaystyle \frac{du}{u} = ln |u| + C\)

Final answer:

\(\displaystyle \frac{1}{2}ln |4 - x^{2}| + C\)

So why is the fraction (in the final answer) postive as opposed to negative?

 

[MarkFL]

The final result should indeed be:

\(\displaystyle \int\dfrac{x}{4-x^2}\,dx=-\dfrac{1}{2}\ln|4-x^2|+C\)

You may verify this via differentiation:

\(\displaystyle \dfrac{d}{dx}\left(-\dfrac{1}{2}\ln|4-x^2|+C \right)=-\dfrac{1}{2}\cdot\dfrac{-2x}{4-x^2}=\dfrac{x}{4-x^2}\)

 

[Jason76]

The final result should indeed be:

\(\displaystyle \int\dfrac{x}{4-x^2}\,dx=-\dfrac{1}{2}\ln|4-x^2|+C\)

You may verify this via differentiation:

\(\displaystyle \dfrac{d}{dx}\left(-\dfrac{1}{2}\ln|4-x^2|+C \right)=-\dfrac{1}{2}\cdot\dfrac{-2x}{4-x^2}=\dfrac{x}{4-x^2}\)

Yeah, I know what you mean. I don't know why the book says it's positive.

 

[MarkFL]

I would guess a typo. We may integrate as follows:

Let \(\displaystyle u=4-x^2\,\therefore\,du=-2x\) and we have:

\(\displaystyle -\frac{1}{2}\int\frac{1}{u}\,du=-\frac{1}{2}\ln|u|+C=-\frac{1}{2}\ln|4-x^2|+C\)

 

[daon2]

There is a way to remove the leading negative sign, but not to obtain your book's solution

\(\displaystyle -\frac{1}{2}\ln|4-x^2| = \frac{1}{2}\ln\frac{1}{|4-x^2|}\)