What is the integral from 4 to 9 of x(2x^2+1)^6 ?

The integral from 4 to 9 of x(2x^2+1)^6 is (163^7-33^7)/(28)

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integral from 4 to 9 of x(2x^2+1)^6

\int_{4}^{9} x (2 x^{2} + 1)^{6} d x

\frac{16 3 ^{7} - 3 3 ^{7}}{28}

Decimal

1.09182 E 14

Solution steps

\int_{4}^{9} x (2 x^{2} + 1)^{6} d x

Apply u-substitution

= \int_{16}^{81} \frac{( 2 u + 1 ) ^{6}}{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_{16}^{81} (2 u + 1)^{6} d u

Apply u-substitution

= \frac{1}{2} \cdot \int_{33}^{163} \frac{v ^{6}}{2} d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{33}^{163} v^{6} d v

Apply the Power Rule

= \frac{1}{2} \cdot \frac{1}{2} [\frac{v ^{7}}{7}]_{33}^{163}

Simplify

\frac{1}{2} \cdot \frac{1}{2} [\frac{v ^{7}}{7}]_{33}^{163} : \frac{1}{4} [\frac{v ^{7}}{7}]_{33}^{163}

= \frac{1}{4} [\frac{v ^{7}}{7}]_{33}^{163}

Compute the boundaries:

\frac{16 3 ^{7} - 3 3 ^{7}}{7}

= \frac{1}{4} \cdot \frac{16 3 ^{7} - 3 3 ^{7}}{7}

Simplify

= \frac{16 3 ^{7} - 3 3 ^{7}}{28}

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What is the integral from 4 to 9 of x(2x^2+1)^6 ?
The integral from 4 to 9 of x(2x^2+1)^6 is (163^7-33^7)/(28)