What is the integral of (3-1/(x^4)) ?

The integral of (3-1/(x^4)) is 3x+1/(3x^3)+C

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integral of (3-1/(x^4))

\int (3 - \frac{1}{x ^{4}}) d x

3 x + \frac{1}{3 x ^{3}} + C

Solution steps

\int 3 - \frac{1}{x ^{4}} 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 3 d x - \int \frac{1}{x ^{4}} d x

\int 3 d x = 3 x

\int \frac{1}{x ^{4}} d x = - \frac{1}{3 x ^{3}}

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

Simplify

= 3 x + \frac{1}{3 x ^{3}}

Add a constant to the solution

= 3 x + \frac{1}{3 x ^{3}} + C

\frac{d}{d x} (cos (2 x) d x)

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

\frac{11 - x}{7} + \frac{36 + x ^{2}}{3}

\int_{0}^{9} ∣ 2 x - 9 ∣ d x

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

What is the integral of (3-1/(x^4)) ?
The integral of (3-1/(x^4)) is 3x+1/(3x^3)+C