What is the integral of (2x+6)^3e^{x+3} ?

The integral of (2x+6)^3e^{x+3} is 8e^{x+3}((x+3)^3-3(x^2+4x+5))+C

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integral of (2x+6)^3e^{x+3}

\int (2 x + 6)^{3} e^{x + 3} d x

8 e^{x + 3} ((x + 3)^{3} - 3 (x^{2} + 4 x + 5)) + C

Solution steps

\int (2 x + 6)^{3} e^{x + 3} d x

Apply u-substitution

: \int \frac{e ^{\frac{u}{2}} u ^{3}}{2} d u

= \int \frac{e ^{\frac{u}{2}} u ^{3}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int e^{\frac{u}{2}} u^{3} d u

Apply Integration By Parts

: 2 e^{\frac{u}{2}} u^{3} - \int 6 e^{\frac{u}{2}} u^{2} d u

= \frac{1}{2} (2 e^{\frac{u}{2}} u^{3} - \int 6 e^{\frac{u}{2}} u^{2} d u)

\int 6 e^{\frac{u}{2}} u^{2} d u = 6 (2 e^{\frac{u}{2}} u^{2} - 16 (\frac{1}{2} e^{\frac{u}{2}} u - e^{\frac{u}{2}}))

= \frac{1}{2} (2 e^{\frac{u}{2}} u^{3} - 6 (2 e^{\frac{u}{2}} u^{2} - 16 (\frac{1}{2} e^{\frac{u}{2}} u - e^{\frac{u}{2}})))

Substitute back

u = 2 x + 6

= \frac{1}{2} (2 e^{\frac{2 x + 6}{2}} (2 x + 6)^{3} - 6 (2 e^{\frac{2 x + 6}{2}} (2 x + 6)^{2} - 16 (\frac{1}{2} e^{\frac{2 x + 6}{2}} (2 x + 6) - e^{\frac{2 x + 6}{2}})))

Simplify

\frac{1}{2} (2 e^{\frac{2 x + 6}{2}} (2 x + 6)^{3} - 6 (2 e^{\frac{2 x + 6}{2}} (2 x + 6)^{2} - 16 (\frac{1}{2} e^{\frac{2 x + 6}{2}} (2 x + 6) - e^{\frac{2 x + 6}{2}}))) : 8 e^{x + 3} ((x + 3)^{3} - 3 (x^{2} + 4 x + 5))

= 8 e^{x + 3} ((x + 3)^{3} - 3 (x^{2} + 4 x + 5))

Add a constant to the solution

= 8 e^{x + 3} ((x + 3)^{3} - 3 (x^{2} + 4 x + 5)) + C

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What is the integral of (2x+6)^3e^{x+3} ?
The integral of (2x+6)^3e^{x+3} is 8e^{x+3}((x+3)^3-3(x^2+4x+5))+C