What is f(x) = int xe^x-xsqrt(x^2+2)dxf(x)=∫xex−x√x2+2dx if f(0)=-2 f(0)=−2?

1 Answer
Cem Sentin
Oct 3, 2017

f(x)f(x)=int xe^x*dx∫xex⋅dx-int x*sqrt(x^2+2)*dx∫x⋅√x2+2⋅dx

=x*e^xx⋅ex-int e^x*dx∫ex⋅dx-1/3*(x^2+2)^(3/2)+C13⋅(x2+2)32+C

=(x-1)e^x(x−1)ex-1/3*(x^2+2)^(3/2)+C13⋅(x2+2)32+C

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

(0-1)*e^0-1/3*2^(3/2)+C=-2(0−1)⋅e0−13⋅232+C=−2

C=(2sqrt(2))/3-1C=2√23−1

Hence f(x)f(x)=(x-1)e^x(x−1)ex-1/3*(x^2+2)^(3/2)+(2sqrt(2))/3-113⋅(x2+2)32+2√23−1

Explanation:

1) I took integral right side.

2) I imposed f(0)=-2f(0)=−2 condition for finding CC.

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