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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
MCQ
Choose the correct option from given four options : If $\int\frac{\text{dx}}{(\text{x}+2)(\text{x}^2+1)}=\text{a}\log|1+\text{x}^2|+\text{b}\tan^{-1}\text{x}+\frac{1}{5}\log|\text{x}+2|+\text{C},$ then :
  • A
    $\text{a}=\frac{-1}{10},\text{b}=\frac{-2}{5}$
  • B
    $\text{a}=\frac{-1}{10},\text{b}=-\frac{2}{5}$
  • ✓
    $\text{a}=\frac{-1}{10},\text{b}=\frac{2}{5}$
  • D
    $\text{a}=\frac{1}{10},\text{b}=\frac{2}{5}$
✓

Answer

Correct option: C.
$\text{a}=\frac{-1}{10},\text{b}=\frac{2}{5}$
Given that, $\int\frac{\text{dx}}{(\text{x}+2)(\text{x}^2+1)}=\text{a}\log|1+\text{x}^2|+\text{b}\tan^{-1}\text{x}+\frac{1}{5}\log|\text{x}+2|+\text{C}$
Now, $\text{I}=\int\frac{\text{dx}}{(\text{x}+2)(\text{x}^2+1)}$
$\frac{1}{(\text{x}+2)(\text{x}^2+1)}=\frac{\text{A}}{\text{x}+2}+\frac{\text{Bx}+\text{C}}{\text{x}^2+1}$
$\Rightarrow1=\text{A}(\text{x}^2+1)+(\text{Bx}+\text{C})(\text{x}+2)$
$\Rightarrow1=\text{Ax}^2+\text{A}+\text{Bx}^2+2\text{Bx}+\text{Cx}+2\text{C}$
$\Rightarrow1=(\text{A}+\text{B})\text{x}^2+(2\text{B}+\text{C})\text{x}+\text{A}+2\text{C}$
$\Rightarrow\text{A}+\text{B}=0,\text{A}+2\text{C}=1,2\text{B}+\text{C}=0$
We have, $\text{A}=\frac{1}{5},\text{B}=-\frac{1}{5}$ and $\text{C}=\frac{2}{5}$
$\therefore\ \int\frac{\text{dx}}{(\text{x}+2)(\text{x}^2+1)}=\frac{1}{5}\int\frac{1}{\text{x}+2}\text{dx}+\int\frac{-\frac{1}{5}\text{x}+\frac{2}{5}}{\text{x}^2+1}\text{dx}$
$=\frac{1}{5}\int\frac{1}{\text{x}+2}\text{dx}-\frac{1}{5}\int\frac{\text{x}}{1+\text{x}^2}\text{dx}+\frac{1}{5}\int\frac{2}{1+\text{x}^2}\text{dx}$
$=\frac{1}{5}\log|\text{x}+2|-\frac{1}{10}\log|1+\text{x}^2|+\frac{2}{5}\tan^{-1}\text{x}+\text{C}$
$\therefore\ \text{b}=\frac{2}{5}$ and $\text{a}=\frac{-1}{10}$

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