What is the integral of (3x^2-1/(x^2)) ?

The integral of (3x^2-1/(x^2)) is x^3+1/x+C

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integral of (3x^2-1/(x^2))

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

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

Solution steps

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

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

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

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

Simplify

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

Add a constant to the solution

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

y^{^{'}} (1 + e^{x}) = e^{x - y}

\frac{\partial}{\partial x} (sin (x - y) + (x + y))

\frac{\partial}{\partial x} (4 x + 2 y)

f (x) = 5 tan (x), (\frac{π}{4}, 5)

\frac{d}{d x} (sec (1 + x^{2}))

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