What is the integral of (9x(6x^2+1)^2)/2 ?

The integral of (9x(6x^2+1)^2)/2 is 1/8 (6x^2+1)^3+C

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integral of (9x(6x^2+1)^2)/2

\int \frac{9 x ( 6 x ^{2} + 1 ) ^{2}}{2} d x

\frac{1}{8} (6 x^{2} + 1)^{3} + C

Solution steps

\int \frac{9 x ( 6 x ^{2} + 1 ) ^{2}}{2} d x

Take the constant out:

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

= \frac{9}{2} \cdot \int x (6 x^{2} + 1)^{2} d x

Apply u-substitution

= \frac{9}{2} \cdot \int \frac{u ^{2}}{12} d u

Take the constant out:

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

= \frac{9}{2} \cdot \frac{1}{12} \cdot \int u^{2} d u

Apply the Power Rule

= \frac{9}{2} \cdot \frac{1}{12} \cdot \frac{u ^{3}}{3}

Substitute back

u = 6 x^{2} + 1

= \frac{9}{2} \cdot \frac{1}{12} \cdot \frac{( 6 x ^{2} + 1 ) ^{3}}{3}

Simplify

\frac{9}{2} \cdot \frac{1}{12} \cdot \frac{( 6 x ^{2} + 1 ) ^{3}}{3} : \frac{1}{8} (6 x^{2} + 1)^{3}

= \frac{1}{8} (6 x^{2} + 1)^{3}

Add a constant to the solution

= \frac{1}{8} (6 x^{2} + 1)^{3} + C

What is the integral of (9x(6x^2+1)^2)/2 ?
The integral of (9x(6x^2+1)^2)/2 is 1/8 (6x^2+1)^3+C