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

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

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

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

\frac{1}{4 x} - \frac{1}{2 x} + C

Solution steps

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

Apply Integration By Parts

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

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

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

= \frac{1}{4 x} - \frac{1}{2 x}

Add a constant to the solution

= \frac{1}{4 x} - \frac{1}{2 x} + C

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