What is the integral of (3x^2+7x+1)e^x ?

The integral of (3x^2+7x+1)e^x is 3e^xx^2+e^xx+C

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integral of (3x^2+7x+1)e^x

\int (3 x^{2} + 7 x + 1) e^{x} d x

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

Solution steps

\int (3 x^{2} + 7 x + 1) e^{x} d x

Apply Integration By Parts

: (3 x^{2} + 7 x + 1) e^{x} - \int (6 x + 7) e^{x} d x

= (3 x^{2} + 7 x + 1) e^{x} - \int (6 x + 7) e^{x} d x

\int (6 x + 7) e^{x} d x = 6 e^{x} x + e^{x}

= (3 x^{2} + 7 x + 1) e^{x} - (6 e^{x} x + e^{x})

Simplify

(3 x^{2} + 7 x + 1) e^{x} - (6 e^{x} x + e^{x}) : 3 e^{x} x^{2} + e^{x} x

= 3 e^{x} x^{2} + e^{x} x

Add a constant to the solution

= 3 e^{x} x^{2} + e^{x} x + C

y^{^{''}} + 9 y = t^{2} \cdot e^{3 t} + 6

\int (\frac{6}{x}) d x

\frac{1}{q + k e ^{q}}

\frac{\partial}{\partial y} (4 x^{\frac{y}{z}})

y^{^{''}} + y = 3 e^{- 8 x}

What is the integral of (3x^2+7x+1)e^x ?
The integral of (3x^2+7x+1)e^x is 3e^xx^2+e^xx+C