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

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

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

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

\frac{1}{2} (\frac{x ^{3}}{3} - \frac{1}{x}) + C

Solution steps

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

Simplify

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

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

Simplify

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

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

Expand

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

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

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

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

= \frac{1}{2} (\frac{x ^{3}}{3} - \frac{1}{x})

Add a constant to the solution

= \frac{1}{2} (\frac{x ^{3}}{3} - \frac{1}{x}) + C

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