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

The integral of x^4(x^5-4)^4 is 1/25 (x^5-4)^5+C

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

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

\frac{1}{25} (x^{5} - 4)^{5} + C

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

= \frac{1}{5} \cdot \int u^{4} d u

Apply the Power Rule

= \frac{1}{5} \cdot \frac{u ^{5}}{5}

Substitute back

u = x^{5} - 4

= \frac{1}{5} \cdot \frac{( x ^{5} - 4 ) ^{5}}{5}

Simplify

\frac{1}{5} \cdot \frac{( x ^{5} - 4 ) ^{5}}{5} : \frac{1}{25} (x^{5} - 4)^{5}

= \frac{1}{25} (x^{5} - 4)^{5}

Add a constant to the solution

= \frac{1}{25} (x^{5} - 4)^{5} + C

\int \frac{8}{36 - 64 x ^{2}} d x

\int \frac{2 x}{2} d x

\int \frac{1}{( u ) ( 256 - u )} d u

\int (x^{2} + 4 x) e^{x} d x

\int \frac{1}{0.37 x - 0.03} d x

What is the integral of x^4(x^5-4)^4 ?
The integral of x^4(x^5-4)^4 is 1/25 (x^5-4)^5+C