What is the integral of x^3sqrt(5+x^4) ?

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

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integral of x^3sqrt(5+x^4)

\int x^{3} 5 + x^{4} d x

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

Solution steps

\int x^{3} 5 + x^{4} 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 = 5 + x^{4}

= \frac{1}{4} \cdot \frac{2}{3} (5 + x^{4})^{\frac{3}{2}}

Simplify

\frac{1}{4} \cdot \frac{2}{3} (5 + x^{4})^{\frac{3}{2}} : \frac{1}{6} (5 + x^{4})^{\frac{3}{2}}

= \frac{1}{6} (5 + x^{4})^{\frac{3}{2}}

Add a constant to the solution

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

\frac{d}{d x} (3 + f (x) (sin (x)))

\int - 3 v^{2} d v

\int_{0}^{7} x 3 x + 1 d x

f (x) = x^{3 x}

\frac{d y}{d x} = \frac{y ^{2} - 1}{x ^{2} - 1}, y (7) = 7

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