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

The integral from 0 to 1 of x(x^2+1)^5 is 21/4

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

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

\frac{21}{4}

Decimal

5.25

Solution steps

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

Apply u-substitution

: \int_{1}^{2} \frac{u ^{5}}{2} d u

= \int_{1}^{2} \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_{1}^{2} u^{5} d u

Apply the Power Rule

: [\frac{u ^{6}}{6}]_{1}^{2}

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

Compute the boundaries:

\frac{21}{2}

= \frac{1}{2} \cdot \frac{21}{2}

Simplify

= \frac{21}{4}

\frac{d}{d x} ((\frac{x ^{2} + 2}{3})^{2})

y = 2 x^{3} - x^{2} + 3

\frac{d ^{2}}{d x ^{2}} (x^{2} + 4)

\frac{d y}{d x} = 4 x y

\frac{d}{d x} (\frac{x}{5 + x})

What is the integral from 0 to 1 of x(x^2+1)^5 ?
The integral from 0 to 1 of x(x^2+1)^5 is 21/4