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

The integral of 1/((4y)^2) is 1/(16y)-1/(8y)+C

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

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

\frac{1}{16 y} - \frac{1}{8 y} + C

Solution steps

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

Apply Integration By Parts

: \frac{1}{16 y} - \int - \frac{1}{8 y ^{2}} d y

= \frac{1}{16 y} - \int - \frac{1}{8 y ^{2}} d y

\int - \frac{1}{8 y ^{2}} d y = \frac{1}{8 y}

= \frac{1}{16 y} - \frac{1}{8 y}

Add a constant to the solution

= \frac{1}{16 y} - \frac{1}{8 y} + C

f (x) = 6 sin (4 x)

(0, - 2), (3, 0)

\frac{e ^{1 - s}}{1 - s}

x \to \infty lim (x + \frac{3}{x})

\frac{\partial}{\partial x} (arccot (x))

What is the integral of 1/((4y)^2) ?
The integral of 1/((4y)^2) is 1/(16y)-1/(8y)+C