MCQ
$\int_3^8 {\frac{{2 - 3x}}{{x\sqrt {(1 + x)} }}{\rm{ }}} dx$ is equal to
- ✓$2\log \,\left( {3/2{e^3}} \right)$
- B$\log (3/{e^3})$
- C$4\log (3/{e^3})$
- DNone of these
✓
Answer
Correct option: A.
$2\log \,\left( {3/2{e^3}} \right)$
a
(a) We have $\int_3^8 {\frac{{2 - 3x}}{{x\sqrt {1 + x} }}dx = I} $
(a) We have $\int_3^8 {\frac{{2 - 3x}}{{x\sqrt {1 + x} }}dx = I} $
Put $1 + x = {t^2} \Rightarrow dx = 2t\,dt$
When $x = 3 \to 8,$ then $t = 2 \to 3$
$\therefore$ $I = 2\int_2^3 {\frac{{5 - 3{t^2}}}{{{t^2} - 1}}dt} $;
$I = 2\int_2^3 {\left( {\frac{2}{{{t^2} - 1}} - 3} \right)} \,dt$
$I = 2\left[ {\frac{2}{{2.1}}\log \frac{{t - 1}}{{t + 1}} - 3t} \right]_2^3$;
$I = 2\log \left( {\frac{3}{{2{e^3}}}} \right)$.
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