What is the integral of (x^2-12)^5(2x) ?

The integral of (x^2-12)^5(2x) is 1/6 (x^2-12)^6+C

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integral of (x^2-12)^5(2x)

\int (x^{2} - 12)^{5} (2 x) d x

\frac{1}{6} (x^{2} - 12)^{6} + C

Solution steps

\int (x^{2} - 12)^{5} \cdot 2 x 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} - 12)^{5} x d x

Apply u-substitution

: \int \frac{u ^{5}}{2} d u

= 2 \cdot \int \frac{u ^{5}}{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^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= 2 \cdot \frac{1}{2} \cdot \frac{u ^{6}}{6}

Substitute back

u = x^{2} - 12

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} - 12 ) ^{6}}{6}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} - 12 ) ^{6}}{6} : \frac{1}{6} (x^{2} - 12)^{6}

= \frac{1}{6} (x^{2} - 12)^{6}

Add a constant to the solution

= \frac{1}{6} (x^{2} - 12)^{6} + C

f (t) = (t - 2)^{2}

\frac{\partial}{\partial x} (\frac{1}{x ^{2} + 2 y ^{2} + 1})

\frac{\partial}{\partial x} (2 cos (x))

\int 3 y e^{y^{3}} d y

f (x) = ln (x - 4)

What is the integral of (x^2-12)^5(2x) ?
The integral of (x^2-12)^5(2x) is 1/6 (x^2-12)^6+C