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

The integral of 6x(3x^2+4)^4 is 1/5 (3x^2+4)^5+C

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{u ^{4}}{6} d u

= 6 \cdot \int \frac{u ^{4}}{6} d u

Take the constant out:

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

= 6 \cdot \frac{1}{6} \cdot \int u^{4} d u

Apply the Power Rule

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

= 6 \cdot \frac{1}{6} \cdot \frac{u ^{5}}{5}

Substitute back

u = 3 x^{2} + 4

= 6 \cdot \frac{1}{6} \cdot \frac{( 3 x ^{2} + 4 ) ^{5}}{5}

Simplify

6 \cdot \frac{1}{6} \cdot \frac{( 3 x ^{2} + 4 ) ^{5}}{5} : \frac{1}{5} (3 x^{2} + 4)^{5}

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{( x - 2 ) ^{3}}{x ^{2}})

\frac{\partial}{\partial x} (x^{3} + y^{3} - 3 x y)

f (x) = 7 e^{x} cos (x), at x = 0

x \to 0 lim ((\frac{x - 4}{x + 1})^{x - 2})

What is the integral of 6x(3x^2+4)^4 ?
The integral of 6x(3x^2+4)^4 is 1/5 (3x^2+4)^5+C