What is the integral from 1 to infinity of x/(x^2+1) ?

The integral from 1 to infinity of x/(x^2+1) is diverges

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integral from 1 to infinity of x/(x^2+1)

\int_{1}^{\infty} \frac{x}{x ^{2} + 1} d x

diverges

Solution steps

\int_{1}^{\infty} \frac{x}{x ^{2} + 1} d x

Compute the indefinite integral:

\int \frac{x}{x ^{2} + 1} d x = \frac{1}{2} ln x^{2} + 1 + C

Compute the boundaries:

\int_{1}^{\infty} \frac{x}{x ^{2} + 1} d x =

diverges

= diverges

\frac{\partial}{\partial y} (x y^{2} e^{x^{2}})

x \to - 1 - lim (2 x^{2} - 4 x + 8)

\frac{\partial}{\partial x} (6 x y + 5 yz + x z - 48)

16 y^{^{''}} + 152 y^{^{'}} + 361 y = 0, y (0) = 9, y^{^{'}} (0) = 2

\frac{\partial}{\partial y} (ln (1 + x^{2} + y))

What is the integral from 1 to infinity of x/(x^2+1) ?
The integral from 1 to infinity of x/(x^2+1) is diverges