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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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Integrate : $\int\large\frac{\log _e \tan ^{-1} (3x)}{4 (1+gx^2)}$

$(a)\;\tan ^{-1} 3x \qquad(b)\;\tan ^{-1} 3x. \log (\tan ^{-1} 3x) \qquad(c)\;\frac{1}{12} \{\tan ^{-1} 3x. \log (\tan ^{-1} 3x)-\tan ^{-1} 3x\} \qquad (d)\;None$

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1 Answer

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$\tan ^{-1} (3x) =t$
$3 \times \large\frac{1}{1+gx^2}$$dx=dt$
=> $ \int \large\frac{\log _e (t)}{4 \times 3}.dt$
=> $\large\frac{1}{12}$$ \int 1. \log _e t .dt$
=> $\large\frac{1}{12}$$ \{\log _e t .t -\int \large\frac{1}{t}$$.t.dt\}$
=> $\large\frac{1}{12} $$\{ t .\log t -t \}+c$
$\large\frac{1}{12}$$ \{\tan ^{-1} 3x. \log (\tan ^{-1} 3x)-\tan ^{-1} 3x\}$
Hence c is the correct answer.

 

answered Dec 23, 2013 by meena.p
edited Dec 23, 2013 by meena.p
 
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