What is the integral of (100-5/(x^2)) ?

The integral of (100-5/(x^2)) is 100x+5/x+C

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integral of (100-5/(x^2))

\int (100 - \frac{5}{x ^{2}}) d x

100 x + \frac{5}{x} + C

Solution steps

\int 100 - \frac{5}{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 100 d x - \int \frac{5}{x ^{2}} d x

\int 100 d x = 100 x

\int \frac{5}{x ^{2}} d x = - \frac{5}{x}

= 100 x - (- \frac{5}{x})

Simplify

= 100 x + \frac{5}{x}

Add a constant to the solution

= 100 x + \frac{5}{x} + C

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What is the integral of (100-5/(x^2)) ?
The integral of (100-5/(x^2)) is 100x+5/x+C