What is the integral of 3x(3-x^{-8}) ?

The integral of 3x(3-x^{-8}) is 3((3x^2)/2+1/(6x^6))+C

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integral of 3x(3-x^{-8})

\int 3 x (3 - x^{- 8}) d x

3 (\frac{3 x ^{2}}{2} + \frac{1}{6 x ^{6}}) + C

Solution steps

\int 3 x (3 - x^{- 8}) d x

Take the constant out:

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

= 3 \cdot \int x (3 - x^{- 8}) d x

Expand

x (3 - x^{- 8}) : 3 x - \frac{1}{x ^{7}}

= 3 \cdot \int 3 x - \frac{1}{x ^{7}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 (\int 3 x d x - \int \frac{1}{x ^{7}} d x)

\int 3 x d x = \frac{3 x ^{2}}{2}

\int \frac{1}{x ^{7}} d x = - \frac{1}{6 x ^{6}}

= 3 (\frac{3 x ^{2}}{2} - (- \frac{1}{6 x ^{6}}))

Simplify

= 3 (\frac{3 x ^{2}}{2} + \frac{1}{6 x ^{6}})

Add a constant to the solution

= 3 (\frac{3 x ^{2}}{2} + \frac{1}{6 x ^{6}}) + C

\frac{d}{d x} (6 cos^{4} (3 x + 2))

x^{2} + 1

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 6 y = 5 e^{3 x}

\frac{( 10 s - 3 )}{s ^{2} + 4 s + 29}

What is the integral of 3x(3-x^{-8}) ?
The integral of 3x(3-x^{-8}) is 3((3x^2)/2+1/(6x^6))+C