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Home  >>  CBSE XII  >>  Math  >>  Integrals
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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} $

Can you answer this question?
 
 

1 Answer

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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.

 

answered Feb 8, 2013 by sreemathi.v
 

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