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

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

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

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

\frac{1}{4} (1 + 3 x^{2})^{4} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= 6 \cdot \int \frac{u ^{3}}{6} d u

Take the constant out:

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

= 6 \cdot \frac{1}{6} \cdot \int u^{3} d u

Apply the Power Rule

= 6 \cdot \frac{1}{6} \cdot \frac{u ^{4}}{4}

Substitute back

u = 1 + 3 x^{2}

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

Simplify

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

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

Add a constant to the solution

= \frac{1}{4} (1 + 3 x^{2})^{4} + C

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