What is the integral of (x^2-9)^3(2x) ?

The integral of (x^2-9)^3(2x) is 1/4 (x^2-9)^4+C

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integral of (x^2-9)^3(2x)

\int (x^{2} - 9)^{3} (2 x) d x

\frac{1}{4} (x^{2} - 9)^{4} + C

Solution steps

\int (x^{2} - 9)^{3} \cdot 2 x d x

Take the constant out:

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

= 2 \cdot \int (x^{2} - 9)^{3} x d x

Apply u-substitution

: \int \frac{u ^{3}}{2} d u

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

Apply the Power Rule

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

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

Substitute back

u = x^{2} - 9

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} - 9 ) ^{4}}{4}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} - 9 ) ^{4}}{4} : \frac{1}{4} (x^{2} - 9)^{4}

= \frac{1}{4} (x^{2} - 9)^{4}

Add a constant to the solution

= \frac{1}{4} (x^{2} - 9)^{4} + C

y = 8 x, (16, 32)

\frac{\partial}{\partial x} (x + \frac{1}{x})

\frac{\partial}{\partial x} ((x - 1)^{2} \cdot (y + 4)^{2})

y^{^{''}} + 4 y^{^{'}} + 4 y = e^{- 2 t}

f (x) = x 3 - 2 x^{2}

What is the integral of (x^2-9)^3(2x) ?
The integral of (x^2-9)^3(2x) is 1/4 (x^2-9)^4+C