What is the integral of y^3sqrt(y^4-5) ?

The integral of y^3sqrt(y^4-5) is 1/6 (y^4-5)^{3/2}+C

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integral of y^3sqrt(y^4-5)

\int y^{3} y^{4} - 5 d y

\frac{1}{6} (y^{4} - 5)^{\frac{3}{2}} + C

Solution steps

\int y^{3} y^{4} - 5 d y

Apply u-substitution

= \int \frac{u}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int u d u

Apply the Power Rule

= \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = y^{4} - 5

= \frac{1}{4} \cdot \frac{2}{3} (y^{4} - 5)^{\frac{3}{2}}

Simplify

\frac{1}{4} \cdot \frac{2}{3} (y^{4} - 5)^{\frac{3}{2}} : \frac{1}{6} (y^{4} - 5)^{\frac{3}{2}}

= \frac{1}{6} (y^{4} - 5)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{6} (y^{4} - 5)^{\frac{3}{2}} + C

What is the integral of y^3sqrt(y^4-5) ?
The integral of y^3sqrt(y^4-5) is 1/6 (y^4-5)^{3/2}+C