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

The integral from 1 to 2 of (x^4+1/(4x^4)) is 3011/480

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

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

\frac{3011}{480}

Decimal Notation

6.27291 \dots

Solution steps

\int_{1}^{2} x^{4} + \frac{1}{4 x ^{4}} 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

= \int_{1}^{2} x^{4} d x + \int_{1}^{2} \frac{1}{4 x ^{4}} d x

\int_{1}^{2} x^{4} d x = \frac{31}{5}

\int_{1}^{2} \frac{1}{4 x ^{4}} d x = \frac{7}{96}

= \frac{31}{5} + \frac{7}{96}

Simplify

\frac{31}{5} + \frac{7}{96} : \frac{3011}{480}

= \frac{3011}{480}

x^{4} - 2 x^{4}

\int \frac{x}{x ^{2} + x - 2} d x

\int - 3 x e^{- 3 x} d x

x = 0 \sum \infty \frac{x}{5 ^{x}}

\frac{\partial}{\partial x} (\frac{2 - x}{2})

What is the integral from 1 to 2 of (x^4+1/(4x^4)) ?
The integral from 1 to 2 of (x^4+1/(4x^4)) is 3011/480