What is the integral from 0 to 1 of (x+1)(x^2+2x)^5 ?

The integral from 0 to 1 of (x+1)(x^2+2x)^5 is 243/4

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integral from 0 to 1 of (x+1)(x^2+2x)^5

\int_{0}^{1} (x + 1) (x^{2} + 2 x)^{5} d x

\frac{243}{4}

Decimal

60.75

Solution steps

\int_{0}^{1} (x + 1) (x^{2} + 2 x)^{5} d x

Apply u-substitution

: \int_{0}^{3} \frac{u ^{5}}{2} d u

= \int_{0}^{3} \frac{u ^{5}}{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_{0}^{3} u^{5} d u

Apply the Power Rule

: [\frac{u ^{6}}{6}]_{0}^{3}

= \frac{1}{2} [\frac{u ^{6}}{6}]_{0}^{3}

Compute the boundaries:

\frac{243}{2}

= \frac{1}{2} \cdot \frac{243}{2}

Simplify

= \frac{243}{4}

\int \frac{- 1800 x}{25 - x ^{2}} d x

t^{2} cos (3 t)

x \to 13 lim (x^{4} - 10 x^{2} - 20)

x \to \infty lim (x e^{(\frac{1}{x})})

\frac{\partial}{\partial x} (e^{x} cos (2 x))

What is the integral from 0 to 1 of (x+1)(x^2+2x)^5 ?
The integral from 0 to 1 of (x+1)(x^2+2x)^5 is 243/4