What is the integral of x^3(x^4+16)^4 ?

The integral of x^3(x^4+16)^4 is 1/20 (x^4+16)^5+C

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integral of x^3(x^4+16)^4

\int x^{3} (x^{4} + 16)^{4} d x

\frac{1}{20} (x^{4} + 16)^{5} + C

Solution steps

\int x^{3} (x^{4} + 16)^{4} d x

Apply u-substitution

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

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

Apply the Power Rule

: \frac{u ^{5}}{5}

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

Substitute back

u = x^{4} + 16

= \frac{1}{4} \cdot \frac{( x ^{4} + 16 ) ^{5}}{5}

Simplify

\frac{1}{4} \cdot \frac{( x ^{4} + 16 ) ^{5}}{5} : \frac{1}{20} (x^{4} + 16)^{5}

= \frac{1}{20} (x^{4} + 16)^{5}

Add a constant to the solution

= \frac{1}{20} (x^{4} + 16)^{5} + C

\frac{d}{d x} (- cot (x) csc (x))

x \to 5 lim (2 x^{2} - 3 x + 7)

f (x) = 1 + cos (x)

\frac{x d x + y d y}{( x ^{2} + y ^{2} ) ^{\frac{3}{2}}} = 0

\frac{d ^{2}}{d x ^{2}} (\frac{8}{9 t ^{4}})

What is the integral of x^3(x^4+16)^4 ?
The integral of x^3(x^4+16)^4 is 1/20 (x^4+16)^5+C