What is the integral of 2(2x+8)^4 ?

The integral of 2(2x+8)^4 is 1/5 (2x+8)^5+C

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integral of 2(2x+8)^4

\int 2 (2 x + 8)^{4} d x

\frac{1}{5} (2 x + 8)^{5} + C

Solution steps

\int 2 (2 x + 8)^{4} d x

Take the constant out:

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

= 2 \cdot \int (2 x + 8)^{4} d x

Apply u-substitution

: \int \frac{u ^{4}}{2} d u

= 2 \cdot \int \frac{u ^{4}}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int u^{4} d u

Apply the Power Rule

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

= 2 \cdot \frac{1}{2} \cdot \frac{u ^{5}}{5}

Substitute back

u = 2 x + 8

= 2 \cdot \frac{1}{2} \cdot \frac{( 2 x + 8 ) ^{5}}{5}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( 2 x + 8 ) ^{5}}{5} : \frac{1}{5} (2 x + 8)^{5}

= \frac{1}{5} (2 x + 8)^{5}

Add a constant to the solution

= \frac{1}{5} (2 x + 8)^{5} + C

\int_{0}^{π} 4 x cos (x^{2}) d x

\frac{d x}{d t} = x - 1

\int \frac{1}{( 1 + x ) ^{4}} d x

x \to \frac{π}{2} lim (cot (x))

n = 5 \sum \infty \frac{1}{2 ^{n}}