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I'm having a lot of trouble with this problem, and I suspect my mistake is somewhere in the setup. Here is the problem:

$$\int_C \frac{1}{1+x} ds$$

$$C: r(t) = ti + \frac{2}{3}t^{3/2}j, 0 \le t \le 3.$$

I paramaterize $x$ and $y$ such that $x = t$, and $\displaystyle y = \frac{1}{1+t}$

I then compute the integral from $0$ to $3$ of $<x(t),y(t)> \cdot r'(t)$ with respect to $t$.

The solution in the text is 2, but my own solution keeps approximating around 5.

Any help will be appreciated!

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Can you format your question using MathJax (see FAQ)? It helps readability and potentially solicit more responses. Regards –  Amzoti Apr 5 '13 at 3:51
    
I'll give it a look. Thanks. –  Heath Huffman Apr 5 '13 at 4:18
    
Please make sure I got it correct. Regards –  Amzoti Apr 5 '13 at 4:31

1 Answer 1

$$ds = |r'(t)| \, dt = \sqrt{1+t}$$

So your integral is essentially

$$\int_0^3 dt \: (1+t)^{-1/2}$$

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