What is the integral of x^1((5x^2+2/3)^4)/1 ?

The integral of x^1((5x^2+2/3)^4)/1 is 1/12150 (15x^2+2)^5+C

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integral of x^1((5x^2+2/3)^4)/1

\int x^{1} \frac{( 5 x ^{2} + \frac{2}{3} ) ^{4}}{1} d x

\frac{1}{12150} (15 x^{2} + 2)^{5} + C

Solution steps

\int x^{1} \frac{( 5 x ^{2} + \frac{2}{3} ) ^{4}}{1} d x

Simplify

x^{1} \frac{( 5 x ^{2} + \frac{2}{3} ) ^{4}}{1} : \frac{1}{81} x (15 x^{2} + 2)^{4}

= \int \frac{1}{81} x (15 x^{2} + 2)^{4} d x

Take the constant out:

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

= \frac{1}{81} \cdot \int x (15 x^{2} + 2)^{4} d x

Apply u-substitution

: \int \frac{( 15 u + 2 ) ^{4}}{2} d u

= \frac{1}{81} \cdot \int \frac{( 15 u + 2 ) ^{4}}{2} d u

Take the constant out:

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

= \frac{1}{81} \cdot \frac{1}{2} \cdot \int (15 u + 2)^{4} d u

Apply u-substitution

: \int \frac{v ^{4}}{15} d v

= \frac{1}{81} \cdot \frac{1}{2} \cdot \int \frac{v ^{4}}{15} d v

Take the constant out:

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

= \frac{1}{81} \cdot \frac{1}{2} \cdot \frac{1}{15} \cdot \int v^{4} d v

Apply the Power Rule

: \frac{v ^{5}}{5}

= \frac{1}{81} \cdot \frac{1}{2} \cdot \frac{1}{15} \cdot \frac{v ^{5}}{5}

Substitute back

= \frac{1}{81} \cdot \frac{1}{2} \cdot \frac{1}{15} \cdot \frac{( 15 x ^{2} + 2 ) ^{5}}{5}

Simplify

\frac{1}{81} \cdot \frac{1}{2} \cdot \frac{1}{15} \cdot \frac{( 15 x ^{2} + 2 ) ^{5}}{5} : \frac{1}{12150} (15 x^{2} + 2)^{5}

= \frac{1}{12150} (15 x^{2} + 2)^{5}

Add a constant to the solution

= \frac{1}{12150} (15 x^{2} + 2)^{5} + C

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What is the integral of x^1((5x^2+2/3)^4)/1 ?
The integral of x^1((5x^2+2/3)^4)/1 is 1/12150 (15x^2+2)^5+C