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

The integral from 1 to 3 of sqrt(1+(x^3-1/(4x^3))^2) is 181/9

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

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

\frac{181}{9}

Decimal Notation

20.11111 \dots

Solution steps

\int_{1}^{3} 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_{1}^{3} \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_{1}^{3} \frac{8 x ^{6} + 1 + 16 x ^{12}}{x ^{3}} d x

Apply u-substitution

: \int_{1}^{729} \frac{4 u + 1}{6 u ^{\frac{4}{3}}} d u

= \frac{1}{4} \cdot \int_{1}^{729} \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_{1}^{729} \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_{1}^{729} \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_{1}^{729} \frac{4}{u ^{\frac{1}{3}}} d u + \int_{1}^{729} \frac{1}{u ^{\frac{4}{3}}} d u)

\int_{1}^{729} \frac{4}{u ^{\frac{1}{3}}} d u = 480

\int_{1}^{729} \frac{1}{u ^{\frac{4}{3}}} d u = \frac{8}{3}

= \frac{1}{4} \cdot \frac{1}{6} (480 + \frac{8}{3})

Simplify

\frac{1}{4} \cdot \frac{1}{6} (480 + \frac{8}{3}) : \frac{181}{9}

= \frac{181}{9}

\frac{\partial}{\partial v} (uv)

\frac{\partial}{\partial z} (2 x^{4} y z^{3})

\frac{d}{d x} (\frac{3}{4 x ^{8}})

x y^{^{'}} + y = 6 x y^{2}, y (1) = - 4

1 - 2 sin^{3} (x)

What is the integral from 1 to 3 of sqrt(1+(x^3-1/(4x^3))^2) ?
The integral from 1 to 3 of sqrt(1+(x^3-1/(4x^3))^2) is 181/9