What is the integral of (x^2-2)(x^3-6x)^{207} ?

The integral of (x^2-2)(x^3-6x)^{207} is 1/624 (x^3-6x)^{208}+C

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integral of (x^2-2)(x^3-6x)^{207}

\int (x^{2} - 2) (x^{3} - 6 x)^{207} d x

\frac{1}{624} (x^{3} - 6 x)^{208} + C

Solution steps

\int (x^{2} - 2) (x^{3} - 6 x)^{207} d x

Apply u-substitution

: \int \frac{u ^{207}}{3} d u

= \int \frac{u ^{207}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int u^{207} d u

Apply the Power Rule

: \frac{u ^{208}}{208}

= \frac{1}{3} \cdot \frac{u ^{208}}{208}

Substitute back

u = x^{3} - 6 x

= \frac{1}{3} \cdot \frac{( x ^{3} - 6 x ) ^{208}}{208}

Simplify

\frac{1}{3} \cdot \frac{( x ^{3} - 6 x ) ^{208}}{208} : \frac{1}{624} (x^{3} - 6 x)^{208}

= \frac{1}{624} (x^{3} - 6 x)^{208}

Add a constant to the solution

= \frac{1}{624} (x^{3} - 6 x)^{208} + C

x \to 2 lim (3 x^{2})

\frac{\partial}{\partial x} (- x - 6 y)

y = \frac{1}{x}

\int x^{2} \cdot (1 + x) d x

\frac{d}{d x} (\frac{1}{( 2 x - 1 ) ^{3}})

What is the integral of (x^2-2)(x^3-6x)^{207} ?
The integral of (x^2-2)(x^3-6x)^{207} is 1/624 (x^3-6x)^{208}+C