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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u ^{2}}{2} d u

= \int \frac{u ^{2}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int u^{2} d u

Apply the Power Rule

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

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

Substitute back

u = 4 + x^{4}

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

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d x} = y (3 - y)

x \to + (- 1) - lim (x^{2} + 1)

\frac{d}{d x} (+ 2 x)

f (x) = 6 x, (9, 18)

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