What is #f(x) = int xe^x-xsqrt(x^2+2)dx# if #f(0)=-2 #?

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Answer 1 · 2017-10-03T20:36:26

#f(x)#=#int xe^x*dx#-#int x*sqrt(x^2+2)*dx#

=#x*e^x#-#int e^x*dx#-#1/3*(x^2+2)^(3/2)+C#

=#(x-1)e^x#-#1/3*(x^2+2)^(3/2)+C#

After imposing #f(0)=-2# condition,

#(0-1)*e^0-1/3*2^(3/2)+C=-2#

#C=(2sqrt(2))/3-1#

Hence #f(x)#=#(x-1)e^x#-#1/3*(x^2+2)^(3/2)+(2sqrt(2))/3-1#

1) I took integral right side.

2) I imposed #f(0)=-2# condition for finding #C#.