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[Post New]22 Mar 2008 14:57:20 IST
    Subject: Re:integral
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layman (148)

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Olaaa!! Perrrfect answer. 24  [38 rates]
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\int 2\cdot \tan^{ - 1} x\,dx

Integrating by parts,

=2\cdot x\cdot tan^{ - 1} x - \int \frac {2x}{1 + x^{2}}\,dx

=x\cdot tan^{ - 1} x - \ln{|1 + x^{2}|}

Then applying the limits,

=\frac {\pi}{2} - \ln{2}


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