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

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

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

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

diverges

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

diverges

= diverges

\int_{1}^{\infty} \frac{4}{x ^{5}} d x

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\int_{0}^{\frac{2}{7}} \frac{1}{49 x ^{2} + 4} d x

\int_{1}^{3} e^{- x} d x

\int_{0}^{1} x y x^{2} + y^{2} d y

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