What is the integral from 1 to infinity of x/(6+x^2) ?

The integral from 1 to infinity of x/(6+x^2) is diverges

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integral from 1 to infinity of x/(6+x^2)

\int_{1}^{\infty} \frac{x}{6 + x ^{2}} d x

diverges

Solution steps

\int_{1}^{\infty} \frac{x}{6 + x ^{2}} d x

Compute the indefinite integral:

\int \frac{x}{6 + x ^{2}} d x = \frac{1}{2} ln 6 + x^{2} + C

Compute the boundaries:

\int_{1}^{\infty} \frac{x}{6 + x ^{2}} d x =

diverges

= diverges

\frac{d}{d x} (0.001 x^{3} - 0.14 x^{2} + 7 x + 160)

[(z - 1)^{3} (z + 1)^{2}] - 1

\int \frac{t ^{2} - 3}{- t ^{3} + 9 t + 1} d t

3 x

\frac{d}{d x} (co x)

What is the integral from 1 to infinity of x/(6+x^2) ?
The integral from 1 to infinity of x/(6+x^2) is diverges