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7.1 inderfinite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.1 inderfinite integral1 Mark
MCQ
The value of $\int_{}^{} {\frac{{dx}}{{\sqrt x \,(x + 9)}}dx} $ is equal to
  • A
    ${\tan ^{ - 1}}\sqrt x $
  • B
    ${\tan ^{ - 1}}\left( {\frac{{\sqrt x }}{3}} \right)$
  • C
    $\frac{2}{3}{\tan ^{ - 1}}\sqrt x $
  • ✓
    $\frac{2}{3}{\tan ^{ - 1}}\left( {\frac{{\sqrt x }}{3}} \right)$
✓

Answer

Correct option: D.
$\frac{2}{3}{\tan ^{ - 1}}\left( {\frac{{\sqrt x }}{3}} \right)$
d
(d) We have, $I = \int_{}^{} {\frac{{dx}}{{\sqrt x (x + 9)}}} $
Put $\sqrt x = t$, squaring both sides, we get $x = {t^2}$ and $dx = 2tdt$
$\therefore $$I = 2\int_{}^{} {\frac{{dt}}{{{t^2} + {3^2}}}} = \frac{2}{3}{\tan ^{ - 1}}\left( {\frac{t}{3}} \right)$ ==> $I = \frac{2}{3}{\tan ^{ - 1}}\left( {\frac{{\sqrt x }}{3}} \right)$.

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