What is the integral of 2(x-6)^3 ?

The integral of 2(x-6)^3 is 1/2 (x-6)^4+C

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integral of 2(x-6)^3

\int 2 (x - 6)^{3} d x

\frac{1}{2} (x - 6)^{4} + C

Solution steps

\int 2 (x - 6)^{3} d x

Take the constant out:

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

= 2 \cdot \int (x - 6)^{3} d x

Apply u-substitution

: \int u^{3} d u

= 2 \cdot \int u^{3} d u

Apply the Power Rule

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

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

Substitute back

u = x - 6

= 2 \cdot \frac{( x - 6 ) ^{4}}{4}

Simplify

2 \cdot \frac{( x - 6 ) ^{4}}{4} : \frac{1}{2} (x - 6)^{4}

= \frac{1}{2} (x - 6)^{4}

Add a constant to the solution

= \frac{1}{2} (x - 6)^{4} + C

2 x^{2} cos (x) cot (x)

\int \frac{25}{25 + e ^{x}} d x

x y^{^{'}} + y = - 3 x y^{2}, y (1) = - 4

f (x) = ln^{4} (x)

f (x) = \frac{1}{x - 8}