Evaluating the indefinite integral.

[shamieh]

How does

\(\displaystyle \int \frac{2x + 2}{x^2 + 2x + 5} \, dx\) turn into \(\displaystyle \ln(x^2 + 2x + 5)\)?

How are they getting rid of the numerator are they just dividing by the reciprocal of 2x + 2?

 

[MarkFL]

The simplest thing to do is use the substitution:

\(\displaystyle u=x^2+2x+5\,\therefore\,du=(2x+2)\,dx\)

And now you have:

\(\displaystyle \int\frac{1}{u}\,du\)

 

[shamieh]

Hmm, this was the result of a partial fraction decomposition problem, I guess after solving three formulas a u substitution just seemed to simple (Tauri)(Dance)

 

[juantheron]

\(\displaystyle \int \frac{2x + 2}{x^2 + 2x + 5} \, dx\) turn into \(\displaystyle \ln(x^2 + 2x + 5)\)?

Solution:: Given $\displaystyle \int\frac{2x+2}{x^2+2x+5}dx$

Let $x^2+2x+5 = t$, Then $\left(2x+2\right)dx = dt$

So Integral convert into $\displaystyle \int\frac{1}{t}dt = \ln \left|t\right|+\mathbb{C}$

So $\displaystyle \int\frac{2x+2}{x^2+2x+5}dx = \ln \left|x^2+2x+5\right|+\mathbb{C}$

 

[shamieh]

My teacher has the solution has \(\displaystyle \ln(x^2 + 2x + 5)\) without the abs value signs...Do you think it's a typo on his part or is there any particular reasons I shouldn't put the abs value sign?

 

[MarkFL]

My teacher has the solution has \(\displaystyle \ln(x^2 + 2x + 5)\) without the abs value signs...Do you think it's a typo on his part or is there any particular reasons I shouldn't put the abs value sign?

Complete the square:

\(\displaystyle x^2+2x+5=(x+1)^2+2^2>0\) for all real $x$ so the absolute value signs are not necessary in this case. Your teacher is correct. :D

 

[shamieh]

With that being said, do you think I would get points taken off for just leaving it with the absolute value signs?

 

[MarkFL]

With that being said, do you think I would get points taken off for just leaving it with the absolute value signs?

Well, I can't speak for your professor, but technically either is correct. I don't think I would deduct points personally, however in my mind I would certainly favor the response that shows recognition of the fact that the expression is never negative for any real $x$.

However, let's put aside the notion of point deduction and look instead at the issue of mathematical simplicity and elegance. I think there is something to be said for striving to make our results as simple as possible while still covering all cases. I see putting absolute value signs around an expression that is always non-negative as a form of kludge. I think we should look at such things and remove anything that is unnecessary, so that our results are as lean as possible. (Wink)