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

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

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

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

diverges

Solution steps

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

Compute the indefinite integral:

\int \frac{1}{3 + x} d x = ln ∣ 3 + x ∣ + C

Compute the boundaries:

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

diverges

= diverges

n = 1 \sum \infty \frac{9}{( - 2 ) ^{n}}

x \to 1 - lim (e^{\frac{1}{x - 1}})

\frac{\partial}{\partial x} (8 x y)

y = e^{3 t s i n (2 t)}

\int \frac{1}{( 4 x + 5 )} d x

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