What is the integral of y^3sqrt(2y^4-1) ?

The integral of y^3sqrt(2y^4-1) is 1/12 (2y^4-1)^{3/2}+C

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integral of y^3sqrt(2y^4-1)

\int y^{3} 2 y^{4} - 1 d y

\frac{1}{12} (2 y^{4} - 1)^{\frac{3}{2}} + C

Solution steps

\int y^{3} 2 y^{4} - 1 d y

Apply u-substitution

: \int \frac{u}{8} d u

= \int \frac{u}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int u d u

Apply the Power Rule

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

= \frac{1}{8} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 2 y^{4} - 1

= \frac{1}{8} \cdot \frac{2}{3} (2 y^{4} - 1)^{\frac{3}{2}}

Simplify

\frac{1}{8} \cdot \frac{2}{3} (2 y^{4} - 1)^{\frac{3}{2}} : \frac{1}{12} (2 y^{4} - 1)^{\frac{3}{2}}

= \frac{1}{12} (2 y^{4} - 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{12} (2 y^{4} - 1)^{\frac{3}{2}} + C

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What is the integral of y^3sqrt(2y^4-1) ?
The integral of y^3sqrt(2y^4-1) is 1/12 (2y^4-1)^{3/2}+C