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STD 12 - 7.2 definite integral — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 7.2 definite integral4 Marks
MCQ
$\int\limits_{ - \,1}^1 {\,\,\frac{{{x^3}\, + \,\,|x|\,\, + \,\,1}}{{{x^2}\, + \,\,2\,|x|\,\, + \,\,1}}}\,dx$ $= a\, ln\, 2 + b$ then :
  • A
    $a = 2 ; b = 1$
  • ✓
    $a = 2 ; b = 0$
  • C
    $a = 3 ; b = - 2$
  • D
    $a = 4 ; b = - 1$
✓

Answer

Correct option: B.
$a = 2 ; b = 0$
b
$I =\int\limits_{ - \,1}^1 {\,\,\frac{{{x^3}}}{{{x^2}\,\, + \,\,2\,\left| x \right|\,\, + \,\,1}}} $  $dx +\int\limits_{ - \,1}^1 {\,\,\frac{{\left| x \right|\,\, + \,\,1}}{{{{\left( {\left| x \right|\,\, + \,\,1} \right)}^2}}}} dx $

$\,\Rightarrow 2 \int\limits_0^1 {\,\,\frac{{dx}}{{1\,\, + \,\,x}}} = 2\, ln \,2$
odd $\Rightarrow$ vanishes even 

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