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

The integral from 1 to 6 of (x+1)(x^2+2x)^2 is 36855/2

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

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

\frac{36855}{2}

Decimal

18427.5

Solution steps

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

Apply u-substitution

: \int_{3}^{48} \frac{u ^{2}}{2} d u

= \int_{3}^{48} \frac{u ^{2}}{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_{3}^{48} u^{2} d u

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{3}^{48}

= \frac{1}{2} [\frac{u ^{3}}{3}]_{3}^{48}

Compute the boundaries:

36855

= \frac{1}{2} \cdot 36855

Simplify

= \frac{36855}{2}

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What is the integral from 1 to 6 of (x+1)(x^2+2x)^2 ?
The integral from 1 to 6 of (x+1)(x^2+2x)^2 is 36855/2