Calculus: Early Transcendentals, Chapter 7, 7.1, Section 7.1, Problem 24

int_0^1(x^2+1)e^-xdx
Let's first evaluate the indefinite integral,using the method of integration by parts
int(x^2+1)e^-xdx=(x^2+1)inte^-xdx-int(d/dx(x^2+1)inte^-xdx)dx
=(x^2+1)(-e^-x)-int(2x*(-e^-x))dx
=-(x^2+1)e^-x+2intxe^-xdx
applying again integration by parts,
=-(x^2+1)e^-x+2(x*inte^-xdx-int(d/dx(x)inte^-xdx)dx
=-(x^2+1)e^-x+2(x(-e^-x)-int(-e^-x)dx)
=-(x^2+1)e^-x+2(-xe^-x+inte^-xdx)
=-(x^2+1)e^-x+2(-xe^-x+(-e^-x))
=-(x^2+1)e^-x-2xe^-x-2e^-x
=-e^-x(x^2+1+2x+2)
=-e^-x(x^2+2x+3)
adding a constant to the solution,
=-e^-x(x^2+2x+3)+C
Now evaluate the definite integral,
int_0^1(x^2+1)e^-xdx=[-e^-x(x^2+2x+3)]_0^1
=[-e^-1(1^2+2*1+3)]-[-e^0(0^2+2*0+3)]
=[-e^-1(6)]-[-1*3]
=-6/e+3