What is the integral of 3x(6x^2+1)^3 ?

The integral of 3x(6x^2+1)^3 is 1/16 (6x^2+1)^4+C

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integral of 3x(6x^2+1)^3

\int 3 x (6 x^{2} + 1)^{3} d x

\frac{1}{16} (6 x^{2} + 1)^{4} + C

Solution steps

\int 3 x (6 x^{2} + 1)^{3} d x

Take the constant out:

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

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

Apply u-substitution

= 3 \cdot \int \frac{u ^{3}}{12} d u

Take the constant out:

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

= 3 \cdot \frac{1}{12} \cdot \int u^{3} d u

Apply the Power Rule

= 3 \cdot \frac{1}{12} \cdot \frac{u ^{4}}{4}

Substitute back

u = 6 x^{2} + 1

= 3 \cdot \frac{1}{12} \cdot \frac{( 6 x ^{2} + 1 ) ^{4}}{4}

Simplify

3 \cdot \frac{1}{12} \cdot \frac{( 6 x ^{2} + 1 ) ^{4}}{4} : \frac{1}{16} (6 x^{2} + 1)^{4}

= \frac{1}{16} (6 x^{2} + 1)^{4}

Add a constant to the solution

= \frac{1}{16} (6 x^{2} + 1)^{4} + C

What is the integral of 3x(6x^2+1)^3 ?
The integral of 3x(6x^2+1)^3 is 1/16 (6x^2+1)^4+C