$\int {\cfrac{{dx}}{{x\left( {{x^2} + 1} \right)}}}$ equals

Option a is correct

Let $\cfrac{1}{{x\left( {{x^2} + 1} \right)}} = \cfrac{A}{x} + \cfrac{{Bx + C}}{{{x^2} + 1}}$

$\Rightarrow$ $1 = A\left( {{x^2} + 1} \right) + \left( {Bx + C} \right)\left( x \right)$

….(i)
Putting $x = 0$ in $(i)$,

we get $1 = A\left( {0 + 1} \right)$ $\Rightarrow$ $A = 1$

Comparing coefficients of ${x^2}$ in (i) on both sides ,

we get
$0 = A + B$ $\Rightarrow$ $B = - 1$

Comparing coefficients of $x$ in (i) on both sides ,

we get$C = 0$
$\therefore$ $\int {\cfrac{1}{{x\left( {{x^2} + 1} \right)}}dx = \int {\left[ {\cfrac{1}{x} + \cfrac{{ - x}}{{{x^2} + 1}}} \right]dx} }$

$= \log x - \cfrac{1}{2}\log \left( {{x^2} + 1} \right) + C$

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