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

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

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integral of (4x)/((2x^2-8)^5)

\int \frac{4 x}{( 2 x ^{2} - 8 ) ^{5}} d x

- \frac{1}{4 ( 2 x ^{2} - 8 ) ^{4}} + C

Solution steps

\int \frac{4 x}{( 2 x ^{2} - 8 ) ^{5}} d x

Take the constant out:

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

= 4 \cdot \int \frac{x}{( 2 x ^{2} - 8 ) ^{5}} d x

Apply u-substitution

: \int \frac{1}{4 u ^{5}} d u

= 4 \cdot \int \frac{1}{4 u ^{5}} d u

Take the constant out:

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

= 4 \cdot \frac{1}{4} \cdot \int \frac{1}{u ^{5}} d u

Apply the Power Rule

: - \frac{1}{4 u ^{4}}

= 4 \cdot \frac{1}{4} (- \frac{1}{4 u ^{4}})

Substitute back

u = 2 x^{2} - 8

= 4 \cdot \frac{1}{4} (- \frac{1}{4 ( 2 x ^{2} - 8 ) ^{4}})

Simplify

4 \cdot \frac{1}{4} (- \frac{1}{4 ( 2 x ^{2} - 8 ) ^{4}}) : - \frac{1}{4 ( 2 x ^{2} - 8 ) ^{4}}

= - \frac{1}{4 ( 2 x ^{2} - 8 ) ^{4}}

Add a constant to the solution

= - \frac{1}{4 ( 2 x ^{2} - 8 ) ^{4}} + C

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

y y^{^{'}} = t^{2}, y (0) = 2

y^{^{'}} + 3 y = 1, y (0) = 1

\frac{d y}{d x}, x y = 30

\int_{0}^{4} (y - y + 2) d y

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