How to intergrate this? The answer should be -2 ln(x^2 +1). I thought it was -4x*ln(x^2 +1), can someone explain to me how to get to the answer? It’s not about filling in the 1 and 0, but I didn’t know how to get rid of that :).

Question

#int_0^1 ((-4x)/(x^2 +1))dx#

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Answer 1 · 2018-06-17T09:28:31

the result is given by #-2ln(2)#

Writing
#-2int_0^1 (2x)/(x^2+1)dx#
and this is

#-[-2ln(x^2+1)]_0^1=-2ln(2)#