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

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

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

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (sec (tan^{2} (x^{4})))

n = 0 \sum \infty (\frac{e}{2 π})^{n}

d y = (x y^{2} + 3 x y) d x

y^{^{''}} + 4 y^{^{'}} + 4 y = 25 cos (x)

n = 0 \sum \infty \frac{1}{3 ^{n} - n}

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