$\int \frac{\mathrm{d} x}{x\left(x^2+1\right)}=$

Evaluate the integral $\int \frac{dx}{x(x^2+1)}$ using partial fraction decomposition.

Decompose the integrand by expressing $\frac{1}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1}$
Multiply both sides by $x(x^2+1)$ to obtain $1 = A(x^2+1) + (Bx+C)x = (A+B)x^2 + Cx + A$
Equating coefficients: $A+B = 0$, $C = 0$, and $A = 1$, yielding $A = 1$, $B = -1$, $C = 0$

The decomposition becomes $\frac{1}{x} - \frac{x}{x^2+1}$
Integrating term by term: $\int \frac{1}{x} dx - \int \frac{x}{x^2+1} dx$
The first integral is $\ln|x|$

For the second integral, substitute $u = x^2+1$, $du = 2x dx$ to get $\frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \ln|x^2+1|$

Combining results: $\ln|x| - \frac{1}{2}\ln|x^2+1| + C = \ln\left|\frac{x}{\sqrt{x^2+1}}\right| + C$
For $x > 0$, this simplifies to $\ln\left(\frac{x}{\sqrt{x^2+1}}\right) + C$

The correct answer is A.