Determine:

Question

Determine:

luigi0210 · Answer

$$\int\limits_{0}^{1} \frac{x+1}{x^2+2x} dx$$

zepdrix · Answer

$$\large u=x^2+2x \qquad\to\qquad du=(2x+2)dx\qquad\to\qquad du=2(x+1)dx$$

$$\Large \int\limits_0^1\frac{(x+1)dx}{x^2+2x}$$

zepdrix · Answer

Woops I still needed to divide by 2 to match up our (x+1)dx.
Understand how that works? :o

luigi0210 · Answer

So far, yes.

luigi0210 · Answer

If I did it right, the answer should be DNE

zepdrix · Answer

From that last step, dividing each side by 2 gives us,$$\large \color{purple}{\frac{1}{2}du=(x+1)dx}$$$$\Large \int\limits\limits_0^1\frac{\color{purple}{(x+1)dx}}{x^2+2x} \qquad\to\qquad \int\limits\limits_{x=0}^1\frac{\color{purple}{\dfrac{1}{2}du}}{u}$$

DNE? Hmm let's see..

zepdrix · Answer

$$\large \frac{1}{2}\ln|x^2+2x|\quad|_0^1$$

Oh cause of that pesky 0? :3 I see..

luigi0210 · Answer

Yes, I just wanted to make sure I was right. thanks :)

zepdrix · Answer

Yah it appears this does not converge :O Good job!

luigi0210 · Answer

@Loser66 has something to say

loser66 · Answer

to me, I break the problem into 2 parts: 
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