What is the integral from 0 to 1 of (x^4+8)/x ?

The integral from 0 to 1 of (x^4+8)/x is diverges

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integral from 0 to 1 of (x^4+8)/x

\int_{0}^{1} \frac{x ^{4} + 8}{x} d x

diverges

Solution steps

\int_{0}^{1} \frac{x ^{4} + 8}{x} d x

Expand

\frac{x ^{4} + 8}{x} : x^{3} + \frac{8}{x}

= \int_{0}^{1} x^{3} + \frac{8}{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{8}{x} d x

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

\int_{0}^{1} \frac{8}{x} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int x e^{x^{2} + 1} d x

(1 + 8 x^{2}) (x - x^{2})

d x + e^{4 x} d y = 0

\int \frac{8 ( x ^{2} + 1 )}{4 x ^{2} + 9} d x

What is the integral from 0 to 1 of (x^4+8)/x ?
The integral from 0 to 1 of (x^4+8)/x is diverges