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

The integral from 1 to 2 of (x^3+1/(4x^3)) is 123/32

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

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

\frac{123}{32}

Decimal Notation

3.84375

Solution steps

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

\int_{1}^{2} x^{3} d x = \frac{15}{4}

\int_{1}^{2} \frac{1}{4 x ^{3}} d x = \frac{3}{32}

= \frac{15}{4} + \frac{3}{32}

Simplify

\frac{15}{4} + \frac{3}{32} : \frac{123}{32}

= \frac{123}{32}

\frac{d y}{d x} - \frac{2}{3 x} y + \frac{x ^{2} + 2}{3} y^{4} = 0

\int x \cdot ln (2 x) d x

y^{^{'}} = y (1 - 0.0004 y)

x = 4 y^{2}, x = 8 + 2 y^{2}

\frac{d}{d x} (3 x^{2} e^{x^{3}})

What is the integral from 1 to 2 of (x^3+1/(4x^3)) ?
The integral from 1 to 2 of (x^3+1/(4x^3)) is 123/32