What is the integral of (3-5x^2)^3*(-2x) ?

The integral of (3-5x^2)^3*(-2x) is 1/20 (3-5x^2)^4+C

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integral of (3-5x^2)^3*(-2x)

\int (3 - 5 x^{2})^{3} \cdot (- 2 x) d x

\frac{1}{20} (3 - 5 x^{2})^{4} + C

Solution steps

\int (3 - 5 x^{2})^{3} (- 2 x) d x

Simplify

= \int - 2 x (3 - 5 x^{2})^{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 (3 - 5 x^{2})^{3} d x

Apply u-substitution

: \int - \frac{u ^{3}}{10} d u

= - 2 \cdot \int - \frac{u ^{3}}{10} d u

Take the constant out:

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

= - 2 (- \frac{1}{10} \cdot \int u^{3} d u)

Apply the Power Rule

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

= - 2 (- \frac{1}{10} \cdot \frac{u ^{4}}{4})

Substitute back

u = 3 - 5 x^{2}

= - 2 (- \frac{1}{10} \cdot \frac{( 3 - 5 x ^{2} ) ^{4}}{4})

Simplify

- 2 (- \frac{1}{10} \cdot \frac{( 3 - 5 x ^{2} ) ^{4}}{4}) : \frac{1}{20} (3 - 5 x^{2})^{4}

= \frac{1}{20} (3 - 5 x^{2})^{4}

Add a constant to the solution

= \frac{1}{20} (3 - 5 x^{2})^{4} + C

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What is the integral of (3-5x^2)^3*(-2x) ?
The integral of (3-5x^2)^3*(-2x) is 1/20 (3-5x^2)^4+C