What is the integral of (1-2x^2)^3(-4x) ?

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

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integral of (1-2x^2)^3(-4x)

\int (1 - 2 x^{2})^{3} (- 4 x) d x

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

Solution steps

\int (1 - 2 x^{2})^{3} (- 4 x) d x

Simplify

= \int - 4 x (1 - 2 x^{2})^{3} d x

Take the constant out:

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

= - 4 \cdot \int x (1 - 2 x^{2})^{3} d x

Apply u-substitution

: \int - \frac{u ^{3}}{4} d u

= - 4 \cdot \int - \frac{u ^{3}}{4} d u

Take the constant out:

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

= - 4 (- \frac{1}{4} \cdot \int u^{3} d u)

Apply the Power Rule

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

= - 4 (- \frac{1}{4} \cdot \frac{u ^{4}}{4})

Substitute back

u = 1 - 2 x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\int 250

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