`int_0^1 (x^2 + 1)e^(-x) dx` Evaluate the integral

`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`