What is the integral of (2x-1/(8x))^2 ?

The integral of (2x-1/(8x))^2 is (4x^3)/3-1/2 x-1/(64x)+C

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integral of (2x-1/(8x))^2

\int (2 x - \frac{1}{8 x})^{2} d x

\frac{4 x ^{3}}{3} - \frac{1}{2} x - \frac{1}{64 x} + C

Solution steps

\int (2 x - \frac{1}{8 x})^{2} d x

Expand

(2 x - \frac{1}{8 x})^{2} : 4 x^{2} - \frac{1}{2} + \frac{1}{64 x ^{2}}

= \int 4 x^{2} - \frac{1}{2} + \frac{1}{64 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 4 x^{2} d x - \int \frac{1}{2} d x + \int \frac{1}{64 x ^{2}} d x

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

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

\int \frac{1}{64 x ^{2}} d x = - \frac{1}{64 x}

= \frac{4 x ^{3}}{3} - \frac{1}{2} x - \frac{1}{64 x}

Add a constant to the solution

= \frac{4 x ^{3}}{3} - \frac{1}{2} x - \frac{1}{64 x} + C

\frac{d}{d x} (\frac{x ^{3}}{9} + \frac{3}{4 x})

9 (x^{3} - x), 0

t^{2} e^{5 t}

y = x, y = x^{3}

n = 0 \sum \infty n (\frac{1}{3})^{n}

What is the integral of (2x-1/(8x))^2 ?
The integral of (2x-1/(8x))^2 is (4x^3)/3-1/2 x-1/(64x)+C