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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Integral Calculus

Integrate : $\int \large\frac{3 \tan ^2 [1+\log(x^2)]}{x}$$dx$

\[\begin {array} {1 1} (a)\;\tan(1+ \log x^2 ) -(\log x^2 +1)+c \\ (b)\;\tan(1+ \log x^2 ) +(\log x^2 +1)+c \\ (c)\;\frac{3}{2} \tan(1+ \log x^2 ) -(\log x^2 +1)+c \\ (d)\;None \end {array}\]

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1 Answer

$1+ \log (x^2) =t$
$0+ \large\frac{1}{x^2}$$ \times 2x dx=dt $
=> $ \large\frac{2}{x} $$dx=dt$
=> $ \int \tan ^2 t . \large\frac{3}{2}.$$dt$
=> $\large\frac{3}{2}$$ \int [ \sec^2 t -1].dt$
=> $\large\frac{3}{2}$$ \int [ \tan t -t]+c$
$\large\frac{3}{2}$$ \bigg[\tan(1+ \log x^2 ) -(\log x^2 +1)\bigg]+c$
Hence c is the correct answer.
answered Dec 19, 2013 by meena.p
 

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