Let, I=int(x^2+1)e^-xdxI=∫(x2+1)e−xdx.
Using the following Rule of Integration by Parts (IBP) :
IBP : intuv'dx=uv-intvu'dx.
We take, u=x^2+1, and, v'=e^-x.
:. u'=2x, and, v=inte^-xdx=e^-x/-1=-e^-x.
:. I=-(x^2+1)e^-x-int(-e^-x*2x)dx,
rArr I=-(x^2+1)e^-x+2I_1," where, "I_1=intxe^-xdx.
We once again use IBP for I_1; this time, we choose,
u=x, and, v'=e^-x :. u'=1, and, v=-e^-x.
:. I_1=-xe^-x-int(-e^-x*1)dx,
: I_1=-xe^-x-e^-x.
Utilising I_1 in I, we get,
I=-(x^2+1)e^-x+2{-xe^-x-e^-x},
rArr I=-(x^2+2x+3)e^-x+C.
:. int_0^1(x^2+1)e^-xdx=-[(x^2+2x+3)e^-x]_0^1,
=-[(1+2+3)e^-1-(0+0+3)e^-0].
rArr int_0^1(x^2+1)e^-xdx=-(6/e-3)=3/e(e-2).
Enjoy Maths.!