What is the integral of 5x(9-x^{-9}) ?

The integral of 5x(9-x^{-9}) is 5((9x^2)/2+1/(7x^7))+C

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integral of 5x(9-x^{-9})

\int 5 x (9 - x^{- 9}) d x

5 (\frac{9 x ^{2}}{2} + \frac{1}{7 x ^{7}}) + C

Solution steps

\int 5 x (9 - x^{- 9}) d x

Take the constant out:

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

= 5 \cdot \int x (9 - x^{- 9}) d x

Expand

x (9 - x^{- 9}) : 9 x - \frac{1}{x ^{8}}

= 5 \cdot \int 9 x - \frac{1}{x ^{8}} 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

= 5 (\int 9 x d x - \int \frac{1}{x ^{8}} d x)

\int 9 x d x = \frac{9 x ^{2}}{2}

\int \frac{1}{x ^{8}} d x = - \frac{1}{7 x ^{7}}

= 5 (\frac{9 x ^{2}}{2} - (- \frac{1}{7 x ^{7}}))

Simplify

= 5 (\frac{9 x ^{2}}{2} + \frac{1}{7 x ^{7}})

Add a constant to the solution

= 5 (\frac{9 x ^{2}}{2} + \frac{1}{7 x ^{7}}) + C

What is the integral of 5x(9-x^{-9}) ?
The integral of 5x(9-x^{-9}) is 5((9x^2)/2+1/(7x^7))+C