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

The integral of w^2sqrt(3w^3+1) is 2/27 (3w^3+1)^{3/2}+C

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integral of w^2sqrt(3w^3+1)

\int w^{2} 3 w^{3} + 1 d w

\frac{2}{27} (3 w^{3} + 1)^{\frac{3}{2}} + C

Solution steps

\int w^{2} 3 w^{3} + 1 d w

Apply u-substitution

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

Take the constant out:

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

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

Apply the Power Rule

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

Substitute back

u = 3 w^{3} + 1

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

Simplify

\frac{1}{9} \cdot \frac{2}{3} (3 w^{3} + 1)^{\frac{3}{2}} : \frac{2}{27} (3 w^{3} + 1)^{\frac{3}{2}}

= \frac{2}{27} (3 w^{3} + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{27} (3 w^{3} + 1)^{\frac{3}{2}} + C

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What is the integral of w^2sqrt(3w^3+1) ?
The integral of w^2sqrt(3w^3+1) is 2/27 (3w^3+1)^{3/2}+C