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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= 3 \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

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

Apply the Power Rule

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

Substitute back

u = x^{3} - 4

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{2 x}{e ^{4 x}})

y^{^{'}} = x + y

y^{^{''}} + y = sin (6 t), y (0) = 0, y^{^{'}} (0) = 0

\frac{\partial}{\partial x} (1 + cos^{4} (x^{4}))

sin (ln (w))

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