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

The integral of (x^2+2x+1)e^{7x} is 1/343 (49e^{7x}x^2+84e^{7x}x+37e^{7x})+C

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integral of (x^2+2x+1)e^{7x}

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

\frac{1}{343} (49 e^{7 x} x^{2} + 84 e^{7 x} x + 37 e^{7 x}) + C

Solution steps

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

Apply u-substitution

= \int \frac{1}{7} e^{u} (\frac{u ^{2}}{49} + \frac{2 u}{7} + 1) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{7} \cdot \int e^{u} (\frac{u ^{2}}{49} + \frac{2 u}{7} + 1) d u

Expand

e^{u} (\frac{u ^{2}}{49} + \frac{2 u}{7} + 1) : \frac{e ^{u} u ^{2}}{49} + \frac{2 e ^{u} u}{7} + e^{u}

= \frac{1}{7} \cdot \int \frac{e ^{u} u ^{2}}{49} + \frac{2 e ^{u} u}{7} + e^{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{7} (\int \frac{e ^{u} u ^{2}}{49} d u + \int \frac{2 e ^{u} u}{7} d u + \int e^{u} d u)

\int \frac{e ^{u} u ^{2}}{49} d u = \frac{1}{49} (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

\int \frac{2 e ^{u} u}{7} d u = \frac{2}{7} (e^{u} u - e^{u})

\int e^{u} d u = e^{u}

= \frac{1}{7} (\frac{1}{49} (u^{2} e^{u} - 2 (e^{u} u - e^{u})) + \frac{2}{7} (e^{u} u - e^{u}) + e^{u})

Substitute back

u = 7 x

= \frac{1}{7} (\frac{1}{49} ((7 x)^{2} e^{7 x} - 2 (e^{7 x} \cdot 7 x - e^{7 x})) + \frac{2}{7} (e^{7 x} \cdot 7 x - e^{7 x}) + e^{7 x})

Simplify

\frac{1}{7} (\frac{1}{49} ((7 x)^{2} e^{7 x} - 2 (e^{7 x} \cdot 7 x - e^{7 x})) + \frac{2}{7} (e^{7 x} \cdot 7 x - e^{7 x}) + e^{7 x}) : \frac{1}{343} (49 e^{7 x} x^{2} + 84 e^{7 x} x + 37 e^{7 x})

= \frac{1}{343} (49 e^{7 x} x^{2} + 84 e^{7 x} x + 37 e^{7 x})

Add a constant to the solution

= \frac{1}{343} (49 e^{7 x} x^{2} + 84 e^{7 x} x + 37 e^{7 x}) + C

What is the integral of (x^2+2x+1)e^{7x} ?
The integral of (x^2+2x+1)e^{7x} is 1/343 (49e^{7x}x^2+84e^{7x}x+37e^{7x})+C