$1=A\left(x^{2}+1\right)+(B x+C) x$
Equating the coefficients of $x^{2}, x,$ and constant term, we obtain
$A+B=0$
$C=0$
$A=1$
On solving these equations, we obtain
$A=1, B=-1,$ and $C=0$
$\therefore \frac{1}{x\left(x^{2}+1\right)}=\frac{1}{x}+\frac{-x}{x^{2}+1}$
$\Rightarrow \int \frac{1}{x\left(x^{2}+1\right)} d x=\int\left\{\frac{1}{x}-\frac{x}{x^{2}+1}\right\} d x$
$=\log |x|-\frac{1}{2} \log \left|x^{2}+1\right|+C$
Hence, the correct Answer is $A$.
Alter:
$\Rightarrow \int \frac{1}{x\left(x^{2}+1\right)} d x=\int\left\{\frac{x}{x^{2}\left(x^{2}+1\right)}\right\} d x$
Let $x^{2}=t,$ therefore, $2 x\, d x=d t$
$\therefore \int \frac{x}{x^{2}\left(x^{2}+1\right)} d x=\frac{1}{2} \int \frac{d t}{t(t+1)}=\frac{1}{2} \int \frac{(t+1)-t}{t(t+1)} d t=\frac{1}{2} \int \frac{1}{t}-\frac{1}{t+1} d t$
$=\frac{1}{2}[\log t-\log (t+1)]+C$
$=\log |x|-\frac{1}{2} \log \left|x^{2}+1\right|+C$
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A coin is tossed n times. The probability of geting at least once is greater than 0.8. Then, the least value of n, is: