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

1 Answer
Oct 3, 2017

#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#

Explanation:

1) I took integral right side.

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