What is the integral of sqrt(1+((x^6-1)^2)/(4x^6)) ?

The integral of sqrt(1+((x^6-1)^2)/(4x^6)) is 1/2 ((x^4)/4-1/(2x^2))+C

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integral of sqrt(1+((x^6-1)^2)/(4x^6))

\int 1 + \frac{( x ^{6} - 1 ) ^{2}}{4 x ^{6}} d x

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

Solution steps

\int 1 + \frac{( x ^{6} - 1 ) ^{2}}{4 x ^{6}} d x

Simplify

1 + \frac{( x ^{6} - 1 ) ^{2}}{4 x ^{6}} : \frac{( x ^{2} + 1 ) ( x ^{4} - x ^{2} + 1 )}{2 x ^{3}}

= \int \frac{( x ^{2} + 1 ) ( x ^{4} - x ^{2} + 1 )}{2 x ^{3}} 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 ^{4} - x ^{2} + 1 )}{x ^{3}} d x

Expand

\frac{( x ^{2} + 1 ) ( x ^{4} - x ^{2} + 1 )}{x ^{3}} : x^{3} + \frac{1}{x ^{3}}

= \frac{1}{2} \cdot \int x^{3} + \frac{1}{x ^{3}} 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^{3} d x + \int \frac{1}{x ^{3}} d x)

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

\int \frac{1}{x ^{3}} d x = - \frac{1}{2 x ^{2}}

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

Add a constant to the solution

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

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What is the integral of sqrt(1+((x^6-1)^2)/(4x^6)) ?
The integral of sqrt(1+((x^6-1)^2)/(4x^6)) is 1/2 ((x^4)/4-1/(2x^2))+C