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

The integral of x(3x^2+2)^5 is 1/36 (3x^2+2)^6+C

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integral of x(3x^2+2)^5

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

\frac{1}{36} (3 x^{2} + 2)^{6} + C

Solution steps

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

Apply u-substitution

: \int \frac{( 3 u + 2 ) ^{5}}{2} d u

= \int \frac{( 3 u + 2 ) ^{5}}{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 (3 u + 2)^{5} d u

Apply u-substitution

: \int \frac{v ^{5}}{3} d v

= \frac{1}{2} \cdot \int \frac{v ^{5}}{3} 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}{3} \cdot \int v^{5} d v

Apply the Power Rule

: \frac{v ^{6}}{6}

= \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{v ^{6}}{6}

Substitute back

= \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{( 3 x ^{2} + 2 ) ^{6}}{6}

Simplify

\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{( 3 x ^{2} + 2 ) ^{6}}{6} : \frac{1}{36} (3 x^{2} + 2)^{6}

= \frac{1}{36} (3 x^{2} + 2)^{6}

Add a constant to the solution

= \frac{1}{36} (3 x^{2} + 2)^{6} + C

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What is the integral of x(3x^2+2)^5 ?
The integral of x(3x^2+2)^5 is 1/36 (3x^2+2)^6+C