What is the integral of 2(2x+6)^5 ?

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

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integral of 2(2x+6)^5

\int 2 (2 x + 6)^{5} d x

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

Solution steps

\int 2 (2 x + 6)^{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 + 6)^{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 + 6

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{2 x + 1}{x + 4} d x

\int \frac{12}{x ^{3} + 2 x ^{2} - 3 x} d x

y^{^{''}} - y^{^{'}} - 4 y = 0

\frac{d}{d x} (e^{(x)})

x \to \infty lim (\frac{1}{x} + x)