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\[\int\limits_{-6}^{-3}{1 \over x+2}dx\] Please help with the steps of integration.. I think I can put the -3 and -6 in the equation after you help me integrating it.
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this is a ln function since the numerator is the derivative of the denominator f(x) = x + 2 f'(x) = 1 so \[\int\limits_{-6}^{-1} 1/(x+2) dx = \ln(x + 2)\] so F(x) = ln(x + 2) Now evaluate by F(-3) - F(-6) so ln(-1) - ln(-4)
Thank you!
Denyekwe, if you're still there, can you check my new post?
Look at this image and let me know what you think
It makes quite some sense :)
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What is confusing about it
It's not confusing, but does it work for every thing like that?
Yes as long as it is in that form. Notice that it is just X not (x^something)
Yes. Thank you!
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