What in the integral of (x+1)e^x

∫(x+1)[e^x] dx
Integrate by parts: u' = e^x, v = x + 1
=> u = e^x, v' = 1
∫ (u'*v) dx
= u*v - ∫(u*v') dx
= (x+1)e^x - ∫ (e^x)*1 dx
= (x+1)e^x - e^x + C
= x*(e^x) + C

∫ (x+1)e^x dx
∫ ( x e^(x) + e^x ) dx
∫ x e^(x) dx + ∫ e^x dx

∫ x e^(x) dx <--- by part

u = x. . . . .dv = e^x
du = dx . . . v = e^x

u * v - ∫ v * du
x * e^x - ∫ e^x dx
x * e^x - e^x

x * e^x - e^x + ∫ e^x dx
x * e^x - e^x + e^x + C
x * e^x + C

OR, u can just do:
∫ (x+1)e^x dx
e^x * ( (x+1) - 1 ) + C
e^x * ( x + 1 - 1 ) + C
e^x * ( x ) + C
x * e^x + C

∫ x e^x dx ====> e^x [ x - 1 ] + C
∫ x^2 e^x dx ====> e^x [ x^2 - 2x + 2 ] + C
∫ x^3 e^x dx ====> e^x [ x^3 - 3x^2 + 6x - 6 ] + C
∫ x^4 e^x dx ====> e^x [ x^4 - 4x^3 + 12x^2 - 24x + 24 ] + C

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I Integrate by parts using
u=x+1 => du=dx and dv=e^xdx => v=e^x.
uv-int(v)du
(x+1)*e^x-int(e^x)dx
(x+1)*e^x-e^x+C

int((x + 1)*exp(x), x) = x*exp(x) + C