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

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

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

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

\frac{1}{6} (3 x^{2} + 5)^{2} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{u}{6} d u

= 2 \cdot \int \frac{u}{6} d u

Take the constant out:

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

= 2 \cdot \frac{1}{6} \cdot \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= 2 \cdot \frac{1}{6} \cdot \frac{u ^{2}}{2}

Substitute back

u = 3 x^{2} + 5

= 2 \cdot \frac{1}{6} \cdot \frac{( 3 x ^{2} + 5 ) ^{2}}{2}

Simplify

2 \cdot \frac{1}{6} \cdot \frac{( 3 x ^{2} + 5 ) ^{2}}{2} : \frac{1}{6} (3 x^{2} + 5)^{2}

= \frac{1}{6} (3 x^{2} + 5)^{2}

Add a constant to the solution

= \frac{1}{6} (3 x^{2} + 5)^{2} + C

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