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April 1, 2015

April 1, 2015

Posted by **Darren** on Friday, January 25, 2013 at 3:56pm.

a. 1 + (3/2a)ln[(3-a)/(3+a)]

b. 2a + 3ln[(3-a)/(3+a)]

c. 3/(9-a^2)

d. 1/(2a)

e. None of the above

- Calc -
**Steve**, Friday, January 25, 2013 at 4:00pmx/(x+3) = 1 - 3/(x+3)

Integral is x - 3ln(x+3)

evaluate at a and -1, divide by 2a to get

(a): 1 + (3/2a)ln[(3-a)/(3+a)]

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