# Choose the correct answer in $\Large \int \normalsize\frac{\large dx}{\large x(x^2+1)}$equals

$\begin{array}{1 1}(A)log \mid x \mid-\frac{1}{2}log(x^2+1)+C &(B)log \mid x \mid+\frac{1}{2}log(x^2+1)+C\\(C)-log \mid x \mid+\frac{1}{2}log(x^2+1)+C &(D)\frac{1}{2}log \mid x \mid+log(x^2+1)+C\end{array}$

Toolbox:
• $\;$(i)If the given function is a rational function of the form $\frac{px+q}{(x+a)(x^2+b)}$,then it can be resolved into its partial function as $\frac{A}{(x+a)}+\frac{Bx+c}{(x^2+b)}.$
• $\int\frac{dx}{x}=log |x|+c.$
Given $I=\int\frac{dx}{x(x^2+1)}dx.$

$\frac{1}{x(x^2+1)}$ can be resolved as $\frac{A}{x}+\frac{Bx+C}{x^2+1}.$

Hence $\frac{1}{x(x^2+1)}$ can be resolved as $\frac{A}{x}+\frac{Bx+C}{x^2+1}.$

$\Rightarrow A(x^2+1)+(Bx+C)x=1.$

To find the values of A and B,let us equate the coefficients of like terms.

Equating the coefficients of $x^2$

A+B=0-----(1)

Equating the coefficients of $x$

C=0-----(2)

Equating the constant terms

A=1----(3)

On solving equ(1),equ(2) and equ(3) we get the values of A,B,C

Since A=1,substitute for A in equ(1)

A+B=0$\Rightarrow 1+B=0$

B=-1.

Hence A=1,B=-1 and C=0.

Substitute for A and B we get

$\frac{A}{x}+\frac{bx+c}{x^2+1}=\frac{1}{x}+\frac{-x+0}{x^2+1}.$

This implies integration of I is

$I=\int\big(\frac{1}{x}+\frac{-(x+0)}{x^2+1}\big)dx.$

On separating the terms we get

$I=\int\frac{1}{x}dx-\int\frac{x}{x^2+1}dx.$

On integrating,

Put $x^2+1=t.$

On differentiating we get

2xdx=dt.$\Rightarrow xdx=\frac{dt}{2}.$

Substituting for t and dt we get

$I=\int\frac{1}{x}dx-\int\frac{dt/2}{t}.$

$\;\;\;=\int\frac{1}{x}dx-\frac{1}{2}\int\frac{dt}{t}.$

$\;\;\;=log|x|-\frac{1}{2}log|t|+c.$

Substituting for t we get

$\;\;\;=log|x|-\frac{1}{2}log|x^2+1|+c.$

Hence A is the correct answer.