What is the integral from 0 to 3 of 3x(1+3x^2)^2 ?

The integral from 0 to 3 of 3x(1+3x^2)^2 is 7317/2

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integral from 0 to 3 of 3x(1+3x^2)^2

\int_{0}^{3} 3 x (1 + 3 x^{2})^{2} d x

\frac{7317}{2}

Decimal Notation

3658.5

Solution steps

\int_{0}^{3} 3 x (1 + 3 x^{2})^{2} d x

Take the constant out:

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

= 3 \cdot \int_{0}^{3} x (1 + 3 x^{2})^{2} d x

Apply u-substitution

: \int_{1}^{28} \frac{u ^{2}}{6} d u

= 3 \cdot \int_{1}^{28} \frac{u ^{2}}{6} d u

Take the constant out:

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

= 3 \cdot \frac{1}{6} \cdot \int_{1}^{28} u^{2} d u

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{1}^{28}

= 3 \cdot \frac{1}{6} [\frac{u ^{3}}{3}]_{1}^{28}

Simplify

3 \cdot \frac{1}{6} [\frac{u ^{3}}{3}]_{1}^{28} : \frac{1}{2} [\frac{u ^{3}}{3}]_{1}^{28}

= \frac{1}{2} [\frac{u ^{3}}{3}]_{1}^{28}

Compute the boundaries:

7317

= \frac{1}{2} \cdot 7317

Simplify

= \frac{7317}{2}

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What is the integral from 0 to 3 of 3x(1+3x^2)^2 ?
The integral from 0 to 3 of 3x(1+3x^2)^2 is 7317/2