What is the integral of 2(2x+8)^5 ?

The integral of 2(2x+8)^5 is 1/6 (2x+8)^6+C

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integral of 2(2x+8)^5

\int 2 (2 x + 8)^{5} d x

\frac{1}{6} (2 x + 8)^{6} + C

Solution steps

\int 2 (2 x + 8)^{5} 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)^{5} d x

Apply u-substitution

: \int \frac{u ^{5}}{2} d u

= 2 \cdot \int \frac{u ^{5}}{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^{5} d u

Apply the Power Rule

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

= 2 \cdot \frac{1}{2} \cdot \frac{u ^{6}}{6}

Substitute back

u = 2 x + 8

= 2 \cdot \frac{1}{2} \cdot \frac{( 2 x + 8 ) ^{6}}{6}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( 2 x + 8 ) ^{6}}{6} : \frac{1}{6} (2 x + 8)^{6}

= \frac{1}{6} (2 x + 8)^{6}

Add a constant to the solution

= \frac{1}{6} (2 x + 8)^{6} + C

\frac{\partial}{\partial x} ((x + 2)^{2} (y + 3))

\int_{0}^{6} \frac{1}{36 - x ^{2}} d x

\int (y + 2) d y

x \to 3 lim ((x) + (4 - x))

f (x) = \frac{1 - x}{x}