What is the integral of e^x(x+1)^2 ?

The integral of e^x(x+1)^2 is e^xx^2+e^x+C

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integral of e^x(x+1)^2

\int e^{x} (x + 1)^{2} d x

e^{x} x^{2} + e^{x} + C

Solution steps

\int e^{x} (x + 1)^{2} d x

Apply Integration By Parts

: (x + 1)^{2} e^{x} - \int 2 (x + 1) e^{x} d x

= (x + 1)^{2} e^{x} - \int 2 (x + 1) e^{x} d x

\int 2 (x + 1) e^{x} d x = 2 e^{x} x

= (x + 1)^{2} e^{x} - 2 e^{x} x

Simplify

(x + 1)^{2} e^{x} - 2 e^{x} x : e^{x} x^{2} + e^{x}

= e^{x} x^{2} + e^{x}

Add a constant to the solution

= e^{x} x^{2} + e^{x} + C

y = x^{2} - 4 x, y = 0

23 x^{3} + x^{2} - 12 x + 17

\frac{d y}{d x} = \frac{x - 3}{2 - y}

f (x) = x \cdot ln (x)

y^{^{''}} + 16 y^{^{'}} + 65 y = 6 e^{- 8 t} cos (t)