MCQ
$\int_{}^{} {\frac{{{x^2} + x - 1}}{{{x^2} + x - 6}}\;dx = } $
- A$x + \log (x + 3) + \log (x - 2) + c$
- ✓$x - \log (x + 3) + \log (x - 2) + c$
- C$x - \log (x + 3) - \log (x - 2) + c$
- DNone of these
✓
Answer
Correct option: B.
$x - \log (x + 3) + \log (x - 2) + c$
b
(b)$\int_{}^{} {\frac{{{x^2} + x - 1}}{{{x^2} + x - 6}}\,dx} = \int_{}^{} {\left[ {1 + \frac{5}{{{x^2} + x - 6}}} \right]} \,dx$
$ = \int_{}^{} {\left[ {1 + \frac{5}{{(x + 3)(x - 2)}}} \right]} \,dx$$ = \int_{}^{} {dx} + \int_{}^{} {\frac{{dx}}{{x - 2}}} - \int_{}^{} {\frac{{dx}}{{x + 3}}} $
$ = x + \log (x - 2) - \log (x + 3) + c$.
(b)$\int_{}^{} {\frac{{{x^2} + x - 1}}{{{x^2} + x - 6}}\,dx} = \int_{}^{} {\left[ {1 + \frac{5}{{{x^2} + x - 6}}} \right]} \,dx$
$ = \int_{}^{} {\left[ {1 + \frac{5}{{(x + 3)(x - 2)}}} \right]} \,dx$$ = \int_{}^{} {dx} + \int_{}^{} {\frac{{dx}}{{x - 2}}} - \int_{}^{} {\frac{{dx}}{{x + 3}}} $
$ = x + \log (x - 2) - \log (x + 3) + c$.
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