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

The integral from 0 to 1 of v(u+v^2)^4 is (v(5v^8+10v^6+10v^4+5v^2+1))/5

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integral from 0 to 1 of v(u+v^2)^4

\int_{0}^{1} v (u + v^{2})^{4} d u

\frac{v ( 5 v ^{8} + 10 v ^{6} + 10 v ^{4} + 5 v ^{2} + 1 )}{5}

Solution steps

\int_{0}^{1} v (u + v^{2})^{4} d u

Take the constant out:

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

= v \cdot \int_{0}^{1} (v^{2} + u)^{4} d u

Apply u-substitution

: \int_{v^{2}}^{v^{2} + 1} w^{4} d w

= v \cdot \int_{v^{2}}^{v^{2} + 1} w^{4} d w

Apply the Power Rule

: [\frac{w ^{5}}{5}]_{v^{2}}^{v^{2} + 1}

= v [\frac{w ^{5}}{5}]_{v^{2}}^{v^{2} + 1}

Compute the boundaries:

\frac{( v ^{2} + 1 ) ^{5}}{5} - \frac{v ^{10}}{5}

= v (\frac{( v ^{2} + 1 ) ^{5}}{5} - \frac{v ^{10}}{5})

Simplify

= \frac{v ( 5 v ^{8} + 10 v ^{6} + 10 v ^{4} + 5 v ^{2} + 1 )}{5}

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What is the integral from 0 to 1 of v(u+v^2)^4 ?
The integral from 0 to 1 of v(u+v^2)^4 is (v(5v^8+10v^6+10v^4+5v^2+1))/5