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

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

integral of sqrt(1+y^4)y^3

\int 1 + y^{4} y^{3} d y

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

Solution steps

\int 1 + y^{4} y^{3} 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 = 1 + y^{4}

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

Simplify

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

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

Add a constant to the solution

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

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