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

The integral of sqrt(1+((x^3)/4-1/(x^3))^2) is (x^6-8)/(16x^2)+C

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

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

\frac{x ^{6} - 8}{16 x ^{2}} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{2} + 4}{12 u ^{\frac{5}{3}}} d u

= \int \frac{u ^{2} + 4}{12 u ^{\frac{5}{3}}} d u

Take the constant out:

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

= \frac{1}{12} \cdot \int \frac{u ^{2} + 4}{u ^{\frac{5}{3}}} d u

Expand

\frac{u ^{2} + 4}{u ^{\frac{5}{3}}} : u^{\frac{1}{3}} + \frac{4}{u ^{\frac{5}{3}}}

= \frac{1}{12} \cdot \int u^{\frac{1}{3}} + \frac{4}{u ^{\frac{5}{3}}} d u

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}{12} (\int u^{\frac{1}{3}} d u + \int \frac{4}{u ^{\frac{5}{3}}} d u)

\int u^{\frac{1}{3}} d u = \frac{3}{4} u^{\frac{4}{3}}

\int \frac{4}{u ^{\frac{5}{3}}} d u = - \frac{6}{u ^{\frac{2}{3}}}

= \frac{1}{12} (\frac{3}{4} u^{\frac{4}{3}} - \frac{6}{u ^{\frac{2}{3}}})

Substitute back

u = x^{3}

= \frac{1}{12} (\frac{3}{4} (x^{3})^{\frac{4}{3}} - \frac{6}{( x ^{3} ) ^{\frac{2}{3}}})

Simplify

\frac{1}{12} (\frac{3}{4} (x^{3})^{\frac{4}{3}} - \frac{6}{( x ^{3} ) ^{\frac{2}{3}}}) : \frac{x ^{6} - 8}{16 x ^{2}}

= \frac{x ^{6} - 8}{16 x ^{2}}

Add a constant to the solution

= \frac{x ^{6} - 8}{16 x ^{2}} + C

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What is the integral of sqrt(1+((x^3)/4-1/(x^3))^2) ?
The integral of sqrt(1+((x^3)/4-1/(x^3))^2) is (x^6-8)/(16x^2)+C