The value of dx 3 - 2x - x^2 will be - Vidyadip

MCQ

The value of $\int {\frac{{dx}}{{3 - 2x - {x^2}}}} $ will be

✓
$\frac{1}{4}\log \,\left( {\frac{{3 + x}}{{1 - x}}} \right)$
B
$\frac{1}{3}\log \,\left( {\frac{{3 + x}}{{1 - x}}} \right)$
C
$\frac{1}{2}\log \,\left( {\frac{{3 + x}}{{1 - x}}} \right)$
D
$\log \,\left( {\frac{{1 - x}}{{3 + x}}} \right)$

✓

Answer

Correct option: A.

$\frac{1}{4}\log \,\left( {\frac{{3 + x}}{{1 - x}}} \right)$

a
(a)$\int {\frac{{dx}}{{3 - 2x - {x^2}}} = \int {\frac{{dx}}{{4 - ({x^2} + 2x + 1)}}} } $
$ = \int {\frac{{dx}}{{4 - {{(x + 1)}^2}}}} $$ = \int {\frac{{dt}}{{{{(2)}^2} - {t^2}}}} $
Where $x + 1 = t,\,\,\,\,\,\therefore dx = dt$
$\therefore I = \frac{1}{{2.2}}\log \left( {\frac{{2 + t}}{{2 - t}}} \right)$$ = \frac{1}{4}\log \left( {\frac{{2 + x + 1}}{{2 - x - 1}}} \right)$$ = \frac{1}{4}\log \left( {\frac{{3 + x}}{{1 - x}}} \right).$

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