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

The integral from 1 to 4 of (x^3+1/(4x^3)) is 8175/128

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

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

\frac{8175}{128}

Decimal Notation

63.8671875

Solution steps

\int_{1}^{4} x^{3} + \frac{1}{4 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

= \int_{1}^{4} x^{3} d x + \int_{1}^{4} \frac{1}{4 x ^{3}} d x

\int_{1}^{4} x^{3} d x = \frac{255}{4}

\int_{1}^{4} \frac{1}{4 x ^{3}} d x = \frac{15}{128}

= \frac{255}{4} + \frac{15}{128}

Simplify

\frac{255}{4} + \frac{15}{128} : \frac{8175}{128}

= \frac{8175}{128}

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n = 1 \sum \infty - 1^{n}

\frac{d}{d x} (\frac{2 x ^{3} - 7}{x ^{2}})

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\int (8 cos (x) + 9 sin (x)) d x

What is the integral from 1 to 4 of (x^3+1/(4x^3)) ?
The integral from 1 to 4 of (x^3+1/(4x^3)) is 8175/128