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STD 12 - 7.1 inderfinite integral — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 7.1 inderfinite integral4 Marks
MCQ
The value of $\int {\frac{{\sqrt {({x^2} - {a^2})} }}{x}dx} $ will be
  • ✓
    $\sqrt {({x^2} - {a^2})} \, - a{\tan ^{ - 1}}\left[ {\frac{{\sqrt {({x^2} - {a^2})} }}{a}} \right]$
  • B
    $\sqrt {({x^2} - {a^2})} \, + a{\tan ^{ - 1}}\left[ {\frac{{\sqrt {({x^2} - {a^2})} }}{a}} \right]$
  • C
    $\sqrt {({x^2} - {a^2})} \, + {a^2}{\tan ^{ - 1}}[\sqrt {{x^2} - {a^2}} ]$
  • D
    ${\tan ^{ - 1}}x/a + c$
✓

Answer

Correct option: A.
$\sqrt {({x^2} - {a^2})} \, - a{\tan ^{ - 1}}\left[ {\frac{{\sqrt {({x^2} - {a^2})} }}{a}} \right]$
a
(a) Let $\sqrt {({x^2} - {a^2})} = t$ ==> ${x^2} - {a^2} = {t^2}$ ==> ${x^2} = {a^2} + {t^2}$
$\therefore$ $xdx = tdt$
 $\int {\frac{{\sqrt {({x^2} - {a^2})} }}{x}dx} = \int {\frac{{\sqrt {({x^2} - {a^2})} \,x}}{{{x^2}}}dx} $
==> $I = \int {\frac{t}{{{a^2} + {t^2}}}tdt} $$ = \int {\frac{{{t^2}}}{{{a^2} + {t^2}}}dt} $
==> $I = \int {\left( {1 - \frac{{{a^2}}}{{{a^2} + {t^2}}}} \right)\,dt} $$ = t - {a^2}\frac{1}{a}{\tan ^{ - 1}}\left( {\frac{t}{a}} \right)$
==> $I = \sqrt {({x^2} - {a^2})} \, - a{\tan ^{ - 1}}\left[ {\frac{{\left\{ {\sqrt {({x^2} - {a^2})} } \right\}}}{a}} \right]$.

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