What is the integral from 0 to 1 of (x^4+3)/x ?

The integral from 0 to 1 of (x^4+3)/x is diverges

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integral from 0 to 1 of (x^4+3)/x

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

diverges

Solution steps

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

Expand

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

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

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

y = \frac{1}{( 3 x + 8 ) ^{2}}

x \to - 3 lim (5 x)

- 2 9 s^{2} + 1

x \to 2 - lim (f (x))

\frac{d}{d x} ((x^{2} 7^{x})^{9})

What is the integral from 0 to 1 of (x^4+3)/x ?
The integral from 0 to 1 of (x^4+3)/x is diverges