What is the integral of 1/((2x)^3) ?

The integral of 1/((2x)^3) is 1/(8x^2)-3/(16x^2)+C

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integral of 1/((2x)^3)

\int \frac{1}{( 2 x ) ^{3}} d x

\frac{1}{8 x ^{2}} - \frac{3}{16 x ^{2}} + C

Solution steps

\int \frac{1}{( 2 x ) ^{3}} d x

Apply Integration By Parts

: \frac{1}{8 x ^{2}} - \int - \frac{3}{8 x ^{3}} d x

= \frac{1}{8 x ^{2}} - \int - \frac{3}{8 x ^{3}} d x

\int - \frac{3}{8 x ^{3}} d x = \frac{3}{16 x ^{2}}

= \frac{1}{8 x ^{2}} - \frac{3}{16 x ^{2}}

Add a constant to the solution

= \frac{1}{8 x ^{2}} - \frac{3}{16 x ^{2}} + C

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What is the integral of 1/((2x)^3) ?
The integral of 1/((2x)^3) is 1/(8x^2)-3/(16x^2)+C