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

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

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

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

\frac{123}{32}

Solution steps

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

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

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

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

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

Simplify

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

= \frac{123}{32}

ln (1 + 4 x^{2} + 4 x)

\frac{\partial}{\partial x} (e^{- x} sin (2 y))

\frac{\partial}{\partial y} (e^{x y z^{5}} y z^{5})

\frac{d}{d x} (x + cos (x^{2}))

f (x) = x^{2} + x + 9, at x = 2

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