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

The integral of sqrt(1+((x^2)/4-1/(x^2))^2) is (x^3)/(12)-1/x+C

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integral of sqrt(1+((x^2)/4-1/(x^2))^2)

\int 1 + (\frac{x ^{2}}{4} - \frac{1}{x ^{2}})^{2} d x

\frac{x ^{3}}{12} - \frac{1}{x} + C

Solution steps

\int 1 + (\frac{x ^{2}}{4} - \frac{1}{x ^{2}})^{2} d x

Simplify

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

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

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{( x ^{2} + 2 x + 2 ) ( x ^{2} - 2 x + 2 )}{x ^{2}} d x

Expand

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

= \frac{1}{4} \cdot \int x^{2} + \frac{4}{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

= \frac{1}{4} (\int x^{2} d x + \int \frac{4}{x ^{2}} d x)

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

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

= \frac{1}{4} (\frac{x ^{3}}{3} - \frac{4}{x})

Simplify

\frac{1}{4} (\frac{x ^{3}}{3} - \frac{4}{x}) : \frac{x ^{3}}{12} - \frac{1}{x}

= \frac{x ^{3}}{12} - \frac{1}{x}

Add a constant to the solution

= \frac{x ^{3}}{12} - \frac{1}{x} + C

y^{^{''}} - 4 y^{^{'}} = e^{2 x}

(\frac{h}{2})^{2}

\int_{0}^{4} (64 x - 8 x^{2}) d x

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

\frac{\partial}{\partial y} (x ln (x y - 5))

What is the integral of sqrt(1+((x^2)/4-1/(x^2))^2) ?
The integral of sqrt(1+((x^2)/4-1/(x^2))^2) is (x^3)/(12)-1/x+C