What is the integral of-2x^2(x^3+6)^4 ?

The integral of-2x^2(x^3+6)^4 is -2/15 (x^3+6)^5+C

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integral of-2x^2(x^3+6)^4

\int - 2 x^{2} (x^{3} + 6)^{4} d x

- \frac{2}{15} (x^{3} + 6)^{5} + C

Solution steps

\int - 2 x^{2} (x^{3} + 6)^{4} 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} (x^{3} + 6)^{4} d x

Apply u-substitution

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

= - 2 \cdot \int \frac{u ^{4}}{3} d u

Take the constant out:

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

= - 2 \cdot \frac{1}{3} \cdot \int u^{4} d u

Apply the Power Rule

: \frac{u ^{5}}{5}

= - 2 \cdot \frac{1}{3} \cdot \frac{u ^{5}}{5}

Substitute back

u = x^{3} + 6

= - 2 \cdot \frac{1}{3} \cdot \frac{( x ^{3} + 6 ) ^{5}}{5}

Simplify

- 2 \cdot \frac{1}{3} \cdot \frac{( x ^{3} + 6 ) ^{5}}{5} : - \frac{2}{15} (x^{3} + 6)^{5}

= - \frac{2}{15} (x^{3} + 6)^{5}

Add a constant to the solution

= - \frac{2}{15} (x^{3} + 6)^{5} + C

\int 2 x^{3} + \frac{1}{2} + \frac{1}{4 x ^{3}} d x

\int_{1}^{6} \frac{ln ( x )}{x ^{2}} d x

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

x^{3} + y^{3} = 8 x y - 5

\frac{\partial}{\partial y} (\frac{x y}{5 x + 3 y})

What is the integral of-2x^2(x^3+6)^4 ?
The integral of-2x^2(x^3+6)^4 is -2/15 (x^3+6)^5+C