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7.1 inderfinite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.1 inderfinite integral1 Mark
MCQ
$\int {\frac{{{3^x}}}{{\sqrt {{9^x} - 1} }}\,\,dx = } $
  • ✓
    $\frac{1}{{\log 3}}\log |{3^x} + \sqrt {{9^x} - 1} | + c$
  • B
    $\frac{1}{{\log 3}}\log |{9^x} + \sqrt {{9^x} - 1} \,| + \,c$
  • C
    $\frac{1}{{\log 9}}\log |{3^x} + \sqrt {{9^x} - 1} \,| + \,c$
  • D
    $\frac{1}{{\log 9}}\log |\,{3^x} - \sqrt {{9^x} - 1} |\, + c$
✓

Answer

Correct option: A.
$\frac{1}{{\log 3}}\log |{3^x} + \sqrt {{9^x} - 1} | + c$
a
(a)$I = \int {\frac{{{3^x}}}{{\sqrt {{9^x} - 1} }}\,dx} $$ = \int {\frac{{{3^x}dx}}{{\sqrt {\,{{({3^x})}^2} - \,1} }}} $
Put ${3^x} = t$==> ${3^x}\log 3\,dx = dt$ ==> ${3^x}\,dx = dt/\log 3$
$ \Rightarrow I = \frac{1}{{\log 3}}\int {\frac{{dt}}{{\sqrt {{t^2} - 1} }}} $$ = \frac{1}{{\log 3}}\log \left[ {t + \sqrt {{t^2} - 1} } \right] + c$
$ = \frac{1}{{{{\log }_e}3}}\log \left[ {{3^x} + \sqrt {{9^x} - 1} } \right] + c$.

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