What is the integral from 1 to infinity of 1/(x+1) ?

The integral from 1 to infinity of 1/(x+1) is diverges

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integral from 1 to infinity of 1/(x+1)

\int_{1}^{\infty} \frac{1}{x + 1} d x

diverges

Solution steps

\int_{1}^{\infty} \frac{1}{x + 1} d x

Compute the indefinite integral:

\int \frac{1}{x + 1} d x = ln ∣ x + 1 ∣ + C

Compute the boundaries:

\int_{1}^{\infty} \frac{1}{x + 1} d x =

diverges

= diverges

f (x) = \frac{x ^{4}}{8} + \frac{1}{4 x ^{2}}

y = lo g_{x} (e)

\int (sec (x) tan (x)) d x

x^{3} cos (x)

\int_{0}^{\frac{π}{12}} t sin (4 t) d t

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