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 4 lim (\frac{1}{( 4 - x ) ^{2}})

\frac{d}{d x} (a + b x cos (3 x) + c x sin (3 x))

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

\frac{\partial}{\partial y} (\frac{1}{1 + x ^{2} + y ^{2}})

\frac{d}{d x} (x^{2} x^{3} + a^{3})

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