What is the integral of (x^2-8)^4(2x) ?

The integral of (x^2-8)^4(2x) is 1/5 (x^2-8)^5+C

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integral of (x^2-8)^4(2x)

\int (x^{2} - 8)^{4} (2 x) d x

\frac{1}{5} (x^{2} - 8)^{5} + C

Solution steps

\int (x^{2} - 8)^{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} - 8)^{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} - 8

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} - 8 ) ^{5}}{5}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} - 8 ) ^{5}}{5} : \frac{1}{5} (x^{2} - 8)^{5}

= \frac{1}{5} (x^{2} - 8)^{5}

Add a constant to the solution

= \frac{1}{5} (x^{2} - 8)^{5} + C

\frac{\partial}{\partial x} (\frac{6 x - 7 y}{6 x + 7 y})

y = 2 (1.1)^{t}

y = 12 x, (9, 36)

5 y^{^{''}} + 25 y^{^{'}} + 1700 y = 10 sin (20 t)

x (3 x - 1)

What is the integral of (x^2-8)^4(2x) ?
The integral of (x^2-8)^4(2x) is 1/5 (x^2-8)^5+C