What is the integral of 2x(x^2+12)^4 ?

The integral of 2x(x^2+12)^4 is 1/5 (x^2+12)^5+C

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integral of 2x(x^2+12)^4

\int 2 x (x^{2} + 12)^{4} d x

\frac{1}{5} (x^{2} + 12)^{5} + C

Solution steps

\int 2 x (x^{2} + 12)^{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 (x^{2} + 12)^{4} d x

Apply u-substitution

: \int \frac{u ^{4}}{2} d u

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = x^{2} + 12

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 12 ) ^{5}}{5}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 12 ) ^{5}}{5} : \frac{1}{5} (x^{2} + 12)^{5}

= \frac{1}{5} (x^{2} + 12)^{5}

Add a constant to the solution

= \frac{1}{5} (x^{2} + 12)^{5} + C

f (x) = \frac{2}{x}, at x = 4

\frac{d}{d t} (\frac{1}{t ^{4} + 1})

\int_{0}^{2} (2 - x^{2}) d x

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

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

What is the integral of 2x(x^2+12)^4 ?
The integral of 2x(x^2+12)^4 is 1/5 (x^2+12)^5+C