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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

diverges

Solution steps

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

Expand

\frac{x ^{4} + 1}{x} : x^{3} + \frac{1}{x}

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

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int 1 + sinh^{2} (x) d x

\int 3 cos (x) - 4 sin (x) d x

6 e^{6 x}

y = 5 - x^{2}, y = x^{2} - 3

\int \frac{1}{x 16 - x ^{2}} d x

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