What is the integral of 97x^3-3x^{-3} ?

The integral of 97x^3-3x^{-3} is (97x^4)/4+3/(2x^2)+C

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integral of 97x^3-3x^{-3}

\int 97 x^{3} - 3 x^{- 3} d x

\frac{97 x ^{4}}{4} + \frac{3}{2 x ^{2}} + C

Solution steps

\int 97 x^{3} - 3 x^{- 3} d x

3 x^{- 3} = \frac{3}{x ^{3}}

= \int 97 x^{3} - \frac{3}{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

= \int 97 x^{3} d x - \int \frac{3}{x ^{3}} d x

\int 97 x^{3} d x = \frac{97 x ^{4}}{4}

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

= \frac{97 x ^{4}}{4} - (- \frac{3}{2 x ^{2}})

Simplify

= \frac{97 x ^{4}}{4} + \frac{3}{2 x ^{2}}

Add a constant to the solution

= \frac{97 x ^{4}}{4} + \frac{3}{2 x ^{2}} + C

\frac{1}{s ^{3} + 3 s}

x \to 0 lim (\frac{3}{x} - \frac{3}{e ^{x} - 1})

x \to 0 lim (\frac{1}{x ^{2}} + 2)

\int \frac{cos ( x )}{x} d x

\int_{0}^{6} 9 sin (x + 1) d x

What is the integral of 97x^3-3x^{-3} ?
The integral of 97x^3-3x^{-3} is (97x^4)/4+3/(2x^2)+C