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

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

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integral from 0 to 1 of 4x(x^2+1)^5

\int_{0}^{1} 4 x (x^{2} + 1)^{5} d x

21

Solution steps

\int_{0}^{1} 4 x (x^{2} + 1)^{5} d x

Take the constant out:

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

= 4 \cdot \int_{0}^{1} x (x^{2} + 1)^{5} d x

Apply u-substitution

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

= 4 \cdot \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

= 4 \cdot \frac{1}{2} \cdot \int_{1}^{2} u^{5} d u

Apply the Power Rule

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

= 4 \cdot \frac{1}{2} [\frac{u ^{6}}{6}]_{1}^{2}

Simplify

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

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

Compute the boundaries:

\frac{21}{2}

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

Simplify

= 21

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