What is the integral from-1 to 1 of (x^2+6x)^3(2x+6) ?

The integral from-1 to 1 of (x^2+6x)^3(2x+6) is 444

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integral from-1 to 1 of (x^2+6x)^3(2x+6)

\int_{- 1}^{1} (x^{2} + 6 x)^{3} (2 x + 6) d x

444

Solution steps

\int_{- 1}^{1} (x^{2} + 6 x)^{3} (2 x + 6) d x

Apply u-substitution

: \int_{4}^{8} \frac{u ( u ^{2} - 36 ) ^{3}}{128} d u

= \int_{4}^{8} \frac{u ( u ^{2} - 36 ) ^{3}}{128} d u

Take the constant out:

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

= \frac{1}{128} \cdot \int_{4}^{8} u (u^{2} - 36)^{3} d u

Apply u-substitution

: \int_{- 20}^{28} \frac{v ^{3}}{2} d v

= \frac{1}{128} \cdot \int_{- 20}^{28} \frac{v ^{3}}{2} d v

Take the constant out:

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

= \frac{1}{128} \cdot \frac{1}{2} \cdot \int_{- 20}^{28} v^{3} d v

Apply the Power Rule

: [\frac{v ^{4}}{4}]_{- 20}^{28}

= \frac{1}{128} \cdot \frac{1}{2} [\frac{v ^{4}}{4}]_{- 20}^{28}

Simplify

\frac{1}{128} \cdot \frac{1}{2} [\frac{v ^{4}}{4}]_{- 20}^{28} : \frac{1}{256} [\frac{v ^{4}}{4}]_{- 20}^{28}

= \frac{1}{256} [\frac{v ^{4}}{4}]_{- 20}^{28}

Compute the boundaries:

113664

= \frac{1}{256} \cdot 113664

Simplify

= 444

\int_{1}^{9} \frac{1}{x x ^{2} + 9} d x

\frac{d}{d x} (5 x^{3} (2 x^{2} - 1))

tan (e^{2 t}) + e^{t a n (2 t)}

\frac{1}{1 + 2 x}

x \to 3 lim ((x - 0) \cdot (x - 3))

What is the integral from-1 to 1 of (x^2+6x)^3(2x+6) ?
The integral from-1 to 1 of (x^2+6x)^3(2x+6) is 444