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

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

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

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

\frac{2 x ^{6} - 1}{8 x ^{2}} + C

Solution steps

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

Simplify

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

= \int \frac{8 x ^{6} + 1 + 16 x ^{12}}{4 x ^{3}} 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{8 x ^{6} + 1 + 16 x ^{12}}{x ^{3}} d x

Apply u-substitution

: \int \frac{4 u + 1}{6 u ^{\frac{4}{3}}} d u

= \frac{1}{4} \cdot \int \frac{4 u + 1}{6 u ^{\frac{4}{3}}} d u

Take the constant out:

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

= \frac{1}{4} \cdot \frac{1}{6} \cdot \int \frac{4 u + 1}{u ^{\frac{4}{3}}} d u

Expand

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

= \frac{1}{4} \cdot \frac{1}{6} \cdot \int \frac{4}{u ^{\frac{1}{3}}} + \frac{1}{u ^{\frac{4}{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}{4} \cdot \frac{1}{6} (\int \frac{4}{u ^{\frac{1}{3}}} d u + \int \frac{1}{u ^{\frac{4}{3}}} d u)

\int \frac{4}{u ^{\frac{1}{3}}} d u = 6 u^{\frac{2}{3}}

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

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

Substitute back

u = x^{6}

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

Simplify

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

= \frac{2 x ^{6} - 1}{8 x ^{2}}

Add a constant to the solution

= \frac{2 x ^{6} - 1}{8 x ^{2}} + C

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