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

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

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

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

diverges

Solution steps

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

Expand

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

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

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

x \to 0 + lim ((x)^{x})

x \to 3 lim (- 9)

\frac{d}{d x} (\frac{u}{v})

f (x) = \frac{8}{x ^{2}}

\int_{0}^{80} \frac{317}{903 \cdot 80}

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