What is the integral from 0 to 1 of (x^4+9)/x ?

The integral from 0 to 1 of (x^4+9)/x is diverges

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integral from 0 to 1 of (x^4+9)/x

\int_{0}^{1} \frac{x ^{4} + 9}{x} d x

diverges

Solution steps

\int_{0}^{1} \frac{x ^{4} + 9}{x} d x

Expand

\frac{x ^{4} + 9}{x} : x^{3} + \frac{9}{x}

= \int_{0}^{1} x^{3} + \frac{9}{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{9}{x} d x

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

\int_{0}^{1} \frac{9}{x} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{6}{s} + \frac{1}{s ^{2} + 16}

\frac{d}{d x} (a (e^{b x} + b e^{b x} x))

\frac{\partial}{\partial x} (x (4 - x^{2}))

y^{^{''}} - 4 y^{^{'}} + 4 y = x^{2} e^{x}

y^{^{'}} - \frac{y}{x} = x cos (x)

What is the integral from 0 to 1 of (x^4+9)/x ?
The integral from 0 to 1 of (x^4+9)/x is diverges