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How do you evaluate the integral #int x/(x+1)dx# from [0,10]?

1 Answer

Feb 14, 2017

#x/(x+1) = (x+1-1)/(x+1) = 1-1/(x+1)#

Explanation:

So,

#int_0^10 x/(x+1) dx = int_0^10 (1-1/(x+1)) dx#

# = {:x-ln(x+1)]_0^10#

# = 10-ln11#

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