What is the integral of u^3sqrt(u^4+2) ?

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

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integral of u^3sqrt(u^4+2)

\int u^{3} u^{4} + 2 d u

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

Solution steps

\int u^{3} u^{4} + 2 d u

Apply u-substitution

: \int \frac{v}{4} d v

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

v = u^{4} + 2

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

Simplify

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

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

Add a constant to the solution

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

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