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STD 12 - 7.2 definite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.2 definite integral1 Mark
MCQ
$\int\limits_0^{\frac{1}{2}} {\,\,\frac{1}{{1\,\, - \,\,{x^2}}}\,\,\ell n\,\,\frac{{1\, + \,x}}{{1\, - \,x}}} \,dx$ is equal to :
  • ✓
    $\frac{1}{4}\,\,\ell {n^{2\,}}\,\frac{1}{3}$
  • B
    $\frac{1}{2} ln^2 \,3$
  • C
    $-\frac{1}{4} ln^2\, 3$
  • D
    cannot be evaluated.
✓

Answer

Correct option: A.
$\frac{1}{4}\,\,\ell {n^{2\,}}\,\frac{1}{3}$
a
Put ln $(1 + x) - ln (1 - x) = t \Rightarrow \frac{{d\,x}}{{1\,\, - \,\,{x^2}}} = \frac{1}{2} dt$ $I = \frac{1}{2} \int\limits_0^{\ell n\,3} \, t dt =\frac{1}{4}  ln^2 \,3 =\frac{1}{4} ln^2 \frac{1}{3}$

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