What is the integral of sqrt(1+3x^3)(9x^2) ?

The integral of sqrt(1+3x^3)(9x^2) is 2/3 (1+3x^3)^{3/2}+C

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integral of sqrt(1+3x^3)(9x^2)

\int 1 + 3 x^{3} (9 x^{2}) d x

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

Solution steps

\int 1 + 3 x^{3} \cdot 9 x^{2} d x

Take the constant out:

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

= 9 \cdot \int 1 + 3 x^{3} x^{2} d x

Apply u-substitution

: \int \frac{u}{9} d u

= 9 \cdot \int \frac{u}{9} d u

Take the constant out:

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

= 9 \cdot \frac{1}{9} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

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

Substitute back

u = 1 + 3 x^{3}

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of sqrt(1+3x^3)(9x^2) ?
The integral of sqrt(1+3x^3)(9x^2) is 2/3 (1+3x^3)^{3/2}+C