What is the integral of 9x(9-x^{-4}) ?

The integral of 9x(9-x^{-4}) is 9((9x^2)/2+1/(2x^2))+C

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integral of 9x(9-x^{-4})

\int 9 x (9 - x^{- 4}) d x

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

Solution steps

\int 9 x (9 - x^{- 4}) d x

Take the constant out:

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

= 9 \cdot \int x (9 - x^{- 4}) d x

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= 9 \cdot \int x (9 - \frac{1}{x ^{4}}) d x

Expand

x (9 - \frac{1}{x ^{4}}) : 9 x - \frac{1}{x ^{3}}

= 9 \cdot \int 9 x - \frac{1}{x ^{3}} 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

= 9 (\int 9 x d x - \int \frac{1}{x ^{3}} d x)

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

\int \frac{1}{x ^{3}} d x = - \frac{1}{2 x ^{2}}

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

Simplify

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

Add a constant to the solution

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

x \to 3 - lim (\frac{x - 3}{x ^{2} - 9})

y = (x^{3} - 25 x)^{12}, (5, 0)

\int x sin (d) x d x

x \to \frac{π}{4} lim (sec (x))

What is the integral of 9x(9-x^{-4}) ?
The integral of 9x(9-x^{-4}) is 9((9x^2)/2+1/(2x^2))+C