What is the integral of x^3(x^4-1)^5 ?

The integral of x^3(x^4-1)^5 is 1/24 (x^4-1)^6+C

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integral of x^3(x^4-1)^5

\int x^{3} (x^{4} - 1)^{5} d x

\frac{1}{24} (x^{4} - 1)^{6} + C

Solution steps

\int x^{3} (x^{4} - 1)^{5} d x

Apply u-substitution

: \int \frac{u ^{5}}{4} d u

= \int \frac{u ^{5}}{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^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= \frac{1}{4} \cdot \frac{u ^{6}}{6}

Substitute back

u = x^{4} - 1

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

Simplify

\frac{1}{4} \cdot \frac{( x ^{4} - 1 ) ^{6}}{6} : \frac{1}{24} (x^{4} - 1)^{6}

= \frac{1}{24} (x^{4} - 1)^{6}

Add a constant to the solution

= \frac{1}{24} (x^{4} - 1)^{6} + C

y = 4 e^{x} cos (x), (0, 4)

\frac{\partial}{\partial x} (x \cdot e^{y} + ln (x) \cdot y)

\frac{v}{v + \frac{c}{v}}

\int_{0}^{1} e^{- 5 x^{2}} d x

x \to \infty lim (e^{\frac{1}{2} - \frac{1}{x ^{2}}} - 1)

What is the integral of x^3(x^4-1)^5 ?
The integral of x^3(x^4-1)^5 is 1/24 (x^4-1)^6+C