document.write( "Question 939809: it's actually about calculus but i couldn't find any parts about it..
\n" ); document.write( "i was solving integration of tan^-1(x) by parts :
\n" ); document.write( "consider ∫tan^-1(x)x1 dx.
\n" ); document.write( "so let u=tan^-1(x) and du/dx=1/(1^2+x^2)
\n" ); document.write( " dv/dx=1, so v=x.
\n" ); document.write( "∫tan^-1(x)dx = xtan^-1(x)-∫(1/(1+x^2))(x)dx
\n" ); document.write( " = xtan^-1(x)-∫(x/(1+x^2)dx
\n" ); document.write( " = xtan^-1(x)-(1/2)∫(2x/(1+x^2))dx
\n" ); document.write( " = xtan^-1(x)-(1/2)tan^-1(x^2)+c.
\n" ); document.write( "but the answer is xtan^-1(x)-(1/2)ln(1+(x^2))+c.
\n" ); document.write( " Why???
\n" ); document.write( "

Algebra.Com's Answer #573011 by richard1234(7193)\"\" \"About 
You can put this solution on YOUR website!
Note the integration by parts formula: \r
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\n" ); document.write( "\n" ); document.write( "In this case, since and (you forgot x in the numerator)\r
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\n" ); document.write( "\n" ); document.write( "Integrating using the substitution gives , as desired.
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