What is the integral from 0 to 1 of (x^4+2)/x ?

The integral from 0 to 1 of (x^4+2)/x is diverges

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integral from 0 to 1 of (x^4+2)/x

\int_{0}^{1} \frac{x ^{4} + 2}{x} d x

diverges

Solution steps

\int_{0}^{1} \frac{x ^{4} + 2}{x} d x

Expand

\frac{x ^{4} + 2}{x} : x^{3} + \frac{2}{x}

= \int_{0}^{1} x^{3} + \frac{2}{x} 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_{0}^{1} x^{3} d x + \int_{0}^{1} \frac{2}{x} d x

\int_{0}^{1} x^{3} d x = \frac{1}{4}

\int_{0}^{1} \frac{2}{x} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{s - 1}{s + 1}

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

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

- 2 t

\frac{\partial}{\partial x} (ln (\frac{x + y}{x - y}))

What is the integral from 0 to 1 of (x^4+2)/x ?
The integral from 0 to 1 of (x^4+2)/x is diverges