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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals1 Mark
MCQ
If $\int\frac{1}{(\text{x}+2)(\text{x}^2+1)}\text{ dx}=\text{a}\log|1+\text{x}^2|+\text{b}\tan^{-1}\text{x}+\frac{1}{5}\log|\text{x}+2|+\text{C},$ 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.
$\text{a}=-\frac{1}{10},\text{ b}=\frac{2}{5}$
$\text{I}=\int\frac{1}{(\text{x}+2)(\text{x}^2+1)}\text{ dx}$
Consider,
$\frac{1}{(\text{x}+2)(\text{x}^2+1)}=\frac{\text{A}}{\text{x}+2}+\frac{\text{Bx}+\text{C}}{(\text{x}^2+1)}$
$1=\text{A}(\text{x}^2+1)+(\text{Bx}+\text{C})(\text{x}+2)$
Comaring coefficeints and solving it simultaneously we get
$\text{A}=\frac{1}{5},\text{ B}=-\frac{1}{5},\text{ C}=\frac{2}{5}$
$\text{I}=\int\bigg(\frac{1}{5\text{x}+1}+\frac{\frac{-1}{5}\text{x}+\frac{2}{5}}{\text{x}^2+1}\bigg)\text{dx}$
Integrating we get as,
$\frac{1}{5}\log|\text{x}+2|-\frac{1}{10}\log|\text{x}^2+1|+\frac{2}{5}\tan^{-1}\text{x}+\text{C}$
$\text{a}=-\frac{1}{10},\text{ b}=\frac{2}{5}$

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