What is the integral of 20x(5x^2-3)^3 ?

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

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integral of 20x(5x^2-3)^3

\int 20 x (5 x^{2} - 3)^{3} d x

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

Solution steps

\int 20 x (5 x^{2} - 3)^{3} d x

Take the constant out:

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

= 20 \cdot \int x (5 x^{2} - 3)^{3} d x

Apply u-substitution

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

= 20 \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

= 20 \cdot \frac{1}{10} \cdot \int u^{3} d u

Apply the Power Rule

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

= 20 \cdot \frac{1}{10} \cdot \frac{u ^{4}}{4}

Substitute back

u = 5 x^{2} - 3

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

Simplify

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

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

Add a constant to the solution

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

\int (sin^{2} (x))^{2} d x

\frac{d ^{2} y}{d x ^{2}} + 4 y = 0

(f (x) (2 x))^{'}

x \to - 8 lim (x + 8)

\frac{\partial}{\partial x} (\frac{y}{z} + \frac{z}{x} - \frac{x z}{y})

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