What is the integral of x^3sqrt(x^4-1) ?

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

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integral of x^3sqrt(x^4-1)

\int x^{3} x^{4} - 1 d x

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

Solution steps

\int x^{3} x^{4} - 1 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} - 1

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

Simplify

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

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

Add a constant to the solution

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

(1 - x^{\frac{2}{3}})^{\frac{3}{2}}

\frac{d}{d x} (\frac{3}{2} sin (x))

\int \frac{1 - x}{x} d x

\int 3 (x - 1)^{2} d x

\int 3 x^{2} y + 4 x y^{2} d x

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