What is the integral of x^3sqrt(x^4+6) ?

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

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integral of x^3sqrt(x^4+6)

\int x^{3} x^{4} + 6 d x

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

Solution steps

\int x^{3} x^{4} + 6 d x

Apply u-substitution

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

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

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

Substitute back

u = x^{4} + 6

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

Simplify

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

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

Add a constant to the solution

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

n = 2 \sum \infty \frac{2}{n ^{2} - n}

\frac{d}{d x} (a + \frac{b}{r ( x ) ( x ) ^{2}})^{3}

\int_{0}^{2} \frac{1}{x ^{2} + x - 6} d x

\frac{d}{d x} (e^{- 2 x} cos (3 x))

x \to 11 lim (10)

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