What is the integral of 1/4-(100)/(x^2) ?

The integral of 1/4-(100)/(x^2) is 1/4 x+(100)/x+C

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integral of 1/4-(100)/(x^2)

\int \frac{1}{4} - \frac{100}{x ^{2}} d x

\frac{1}{4} x + \frac{100}{x} + C

Solution steps

\int \frac{1}{4} - \frac{100}{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 \frac{1}{4} d x - \int \frac{100}{x ^{2}} d x

\int \frac{1}{4} d x = \frac{1}{4} x

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

= \frac{1}{4} x - (- \frac{100}{x})

Simplify

= \frac{1}{4} x + \frac{100}{x}

Add a constant to the solution

= \frac{1}{4} x + \frac{100}{x} + C

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

6 e^{- 5 t} \cdot cos (t)

n = 1 \sum \infty 5 (- \frac{1}{5})^{n}

f (x) = 2 x^{2} + x, at x = - 2

x \to 0 + lim (ln (x) \cdot ln (1 + x))

What is the integral of 1/4-(100)/(x^2) ?
The integral of 1/4-(100)/(x^2) is 1/4 x+(100)/x+C