How do you evaluate #int x/sqrt(1+x^2)# from #[-sqrt2,0]#?

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Answer 1 · 2018-03-26T16:34:57

#1-sqrt3#

#int_-sqrt2^0x/sqrt(1+x^2)dx#

#int_-sqrt2^0x(1+x^2)^(-1/2)dx#

by inspection

#=[(1+x^2)^(1/2)]_-sqrt2^0#

#=[(1+x^2)^(1/2)]^0-[(1+x^2)^(1/2)]_-sqrt2#

#=1-(1+2)^(1/2)#

#=1-sqrt3#