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

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

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

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

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

Solution steps

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

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} - 4 ) ^{5}}{5}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} - 4 ) ^{5}}{5} : \frac{1}{5} (x^{2} - 4)^{5}

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

Add a constant to the solution

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

\int \frac{x ^{2}}{( x ^{3} - 1 ) ^{4}} d x

\int 9 sin^{- \frac{3}{2}} (x) cos^{3} (x) d x

x \to 1 lim (x - 8)

\frac{d}{d x} (\frac{x ^{3}}{x ^{2} + 1})

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

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