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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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Integrate: $\int \large\frac{e^{-\tan ^{-1}(a^x)}}{[1+(a^x)^2]}$$a^x.dx$

\[\begin {array} {1 1} (a)\;\frac{1}{\log a}.e^{\tan ^{-1}(a^x)}+c \\ (b)\;e^{\tan ^{-1}(a^x)}+c \\ (c)\;\log a.e^{\tan ^{-1}(a^x)}+c \\ (d)\;None \end {array}\]

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1 Answer

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$\tan ^{-1} (a^x)=t$
$\large\frac{1}{[1+(a^x)^2]}$$a^x. \log a dx =dt$
=> $\int \large\frac{e^{-t}}{\log a}$$dt$
=> $\int \large\frac{-1}{ln a}$$ .e^{-t} (-1) +c$
=>$\frac{1}{\log a}.e^{\tan ^{-1}(a^x)}+c$
Hence a is the correct answer.
answered Dec 19, 2013 by meena.p
 
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