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7.1 inderfinite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.1 inderfinite integral1 Mark
MCQ
$\int_{}^{} {\frac{{x - 1}}{{(x - 3)(x - 2)}}dx = } $
  • A
    $\log (x - 3) - \log (x - 2) + c$
  • ✓
    $\log {(x - 3)^2} - \log (x - 2) + c$
  • C
    $\log (x - 3) + \log (x - 2) + c$
  • D
    $\log {(x - 3)^2} + \log (x - 2) + c$
✓

Answer

Correct option: B.
$\log {(x - 3)^2} - \log (x - 2) + c$
b
(b)$\int_{}^{} {\frac{{x - 1}}{{(x - 3)(x - 2)}}\,dx} $
$ = \int_{}^{} {\frac{{x - 3}}{{(x - 3)(x - 2)}}\,dx + \int_{}^{} {\frac{2}{{(x - 3)(x - 2)}}} } \,dx$
$ = \log \left[ {\frac{{(x - 2){{(x - 3)}^2}}}{{{{(x - 2)}^2}}}} \right] + c = \log \left[ {\frac{{{{(x - 3)}^2}}}{{(x - 2)}}} \right] + c.$
Trick : By inspection, $\frac{d}{{dx}}\left\{ {\log (x - 3) - \log (x - 2)} \right\}$
$ = \frac{1}{{x - 3}} - \frac{1}{{x - 2}} = \frac{1}{{(x - 3)(x - 2)}}$
$ \Rightarrow \frac{d}{{dx}}\left\{ {2\log (x - 3) - \log (x - 2)} \right\}$
$ = \frac{2}{{x - 3}} - \frac{1}{{x - 2}} = \frac{{x - 1}}{{(x - 3)(x - 2)}}$.

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