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

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

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

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

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

Solution steps

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

Simplify

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

y^{^{''}} + 9 y = 6 sin (3 t)

2 sec (x)

\frac{d}{d x} (4^{2 - 2 x})

\int_{0}^{1} 2 x (4 - x^{2}) d x

\frac{\partial}{\partial x} (10 x y e^{- (x^{2} + y^{2})})

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