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

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

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

\int (\frac{1}{4 x ^{2}}) d x

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int \frac{1}{x ^{2}} d x

Apply the Power Rule

: - \frac{1}{x}

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

Simplify

\frac{1}{4} (- \frac{1}{x}) : - \frac{1}{4 x}

= - \frac{1}{4 x}

Add a constant to the solution

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

9 x, \frac{3}{4} x, 70 - x^{2}

g (x) = 3 x^{2} lo g_{3} (x)

\frac{d}{d x} (\frac{x ^{3}}{e ^{2 x}})

\int \frac{4}{15} t^{\frac{5}{2}} + 2 cos (t) + Ct + C d t

y = x^{3} - 3 x + 3, (3, 21)