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Home  >>  CBSE XII  >>  Math  >>  Integrals
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If $\Large \int\frac{dx}{(x+2)(x^2+1)}=\normalsize a\log\mid 1+x^2\mid+b\tan^{-1}x+\frac{1}{5}\log\mid x+2\mid+C$,then\begin{array}{1 1}(A)\;a=\frac{-1}{10},b=\frac{-2}{5} & (B)\;a=\frac{1}{10},b=\frac{-2}{5}\\(C)\;a=\frac{-1}{10},b=\frac{2}{5} & (D)\;a=\frac{1}{10},b=\frac{2}{5}\end{array}

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1 Answer

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Toolbox:
  • $\large\frac{dx}{(x+a)(x+b)}$ can be resolved into partial fraction as $\large\frac{A}{x+a}+\frac{B}{x+b}$
  • $\int \large\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\big(\frac{x}{a}\bigg)$$+c$.
Step 1:
Let $I=\int \large\frac{dx}{(x+2)(x^2+1)}$
$\large\frac{1}{(x+2)(x^2+1)}$ can be resolved into partial fraction as
$\large\frac{1}{(x+2)(x^2+1)}=\large\frac{A}{(x+2)}+\frac{Bx+c}{(x^2+1)}$
$\Rightarrow 1=A(x^2+1)+(x+2)(Bx+C)$
Equating the coefficients of like terms,first let us equate for $x^2$
$0=A+B$-------(1)
Next let us equate for $x$
0=2B+C------(2)
Let us equate for the constant term
1=A+2C------(3)
A+B=0$\Rightarrow A=-B.$
Substituting this in equ(3)
-B+2C=1
Now solving equ (2) and equ (3)
-B+2C=1
2B+C=0
Multiply equ (2) by 2
4B+2C=0
-B+2C=1
_______________
5B=-1
$\Rightarrow B=\frac{-1}{5}\Rightarrow A=\frac{1}{5}$
Step 2:
Substituting for B in equ (2)
$2\big(\frac{-1}{5}\big)+c=0$
Hence $A=\large\frac{1}{5}$,$B=\large\frac{-1}{5}$ and $C=\large\frac{2}{5}$
Now substituting for $A,B$ and $C$ we get
$\large\frac{1}{(x+2)(x^2+1)}=\frac{1/5}{(x+2)}+\frac{\frac{-1}{5}x+\frac{2}{5}}{x^2+1}$
$\quad\quad\quad\quad=\large\frac{1}{5(x+2)}-\frac{x}{5(x^2+1)}+\frac{2}{5(x^2+1)}$
$I=\large\frac{1}{5}\int\frac{dx}{x+2}-\frac{1}{5}\int \frac{x}{x^2+1}dx+\frac{2}{5}\int\frac{dx}{x^2+1}$
Step 3:
$\large\frac{1}{5}\int\frac{dx}{x+2}=\frac{1}{5}\log\mid x+2\mid$
Consider $\large\frac{1}{5}\int \frac{x}{x^2+1}dx$
Put $x^2+1=t\Rightarrow 2xdx=dt\Rightarrow xdx=\large\frac{dt}{2}$
On substituting this we get
$\large\frac{-1}{5}\frac{1}{2}\int \frac{dt}{t}=\frac{-1}{10}\log \mid t\mid$
Substituting for t we get $\frac{-1}{10}\log\mid x^2+1\mid$
$\large\frac{2}{5}\int \frac{dx}{x^2+1}$ is of the form $\large\int\frac{dx}{x^2+a^2}=\frac{1}{a}\normalsize \tan^{-1}\big(\large\frac{x}{a}\big)$
Here a=1
On integrating we get,
$\large\frac{2}{5}\normalsize\tan^{-1}(x)$
On combining the above we get,
$I=\frac{1}{5}\log\mid x+2\mid-\frac{1}{10}\log\mid x^2+1\mid+\large\frac{2}{5}\normalsize\tan^{-1}(x)$
$a=\large\frac{-1}{10}$,$b=\large\frac{2}{5}$
Hence the correct answer is C
answered Apr 24, 2013 by sreemathi.v
 

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