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

The integral of sqrt(1+(x/2-1/(2x))^2) is 1/2 ((x^2)/2+ln|x|)+C

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integral of sqrt(1+(x/2-1/(2x))^2)

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

\frac{1}{2} (\frac{x ^{2}}{2} + ln ∣ x ∣) + C

Solution steps

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

Simplify

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

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

Take the constant out:

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

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

Expand

\frac{x ^{2} + 1}{x} : x + \frac{1}{x}

= \frac{1}{2} \cdot \int x + \frac{1}{x} 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

= \frac{1}{2} (\int x d x + \int \frac{1}{x} d x)

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

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{1}{2} (\frac{x ^{2}}{2} + ln ∣ x ∣)

Add a constant to the solution

= \frac{1}{2} (\frac{x ^{2}}{2} + ln ∣ x ∣) + C

(1 + \frac{1}{x})^{3}

7 t y^{^{'}} + y = t^{2}

2 t^{2} - 3 t - 1

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

\frac{1}{( 4 x + 1 ) ^{4}}

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