What is the integral from 0 to 1 of (x^4+6)/x ?

The integral from 0 to 1 of (x^4+6)/x is diverges

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integral from 0 to 1 of (x^4+6)/x

\int_{0}^{1} \frac{x ^{4} + 6}{x} d x

diverges

Solution steps

\int_{0}^{1} \frac{x ^{4} + 6}{x} d x

Expand

\frac{x ^{4} + 6}{x} : x^{3} + \frac{6}{x}

= \int_{0}^{1} x^{3} + \frac{6}{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{6}{x} d x

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

\int_{0}^{1} \frac{6}{x} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

tan^{4} (2 x - 1)^{3}

\int \frac{x}{x + 1 - x + 1} d x

x \to 1 + lim (\frac{3}{ln ( x )} - \frac{2}{x - 1})

y = \frac{4}{x}, (5, \frac{4}{5})

(ln (7 x))^{2}

What is the integral from 0 to 1 of (x^4+6)/x ?
The integral from 0 to 1 of (x^4+6)/x is diverges