1I=∫(tan-1x)2dx
Integration by parts
I=x(tan-1)2-∫2xtan-1x1+x2dx
I'=∫2xtan-1x1+x2dx
tan-1x=t
dx=(1+x2)dt
I'=2∫t tant dt
I'=2tlog!sect!-2∫log!sect!dt
I'=2tlog!sect!-2tlog!sect!-∫t tantdt
I'=-4tlog!sect!-I'/2
I'=8tlog!sect!/3
I=x(tan-1x)2-8tlog!sect!/3 +C
1I do not use any software.Whatever facility is given on this forum I use it.
1Vinay , in line no. 9 , you get
I ' = 2 t log ( sec t ) - 2 t log ( sec t ) + I ' ........................ : )
You have part integrated taking one function , then again part integrated taking the other function , which leaves you at -
I ' = I '
Actually , this is an un - integrable function . You cannot find its antiderivative in terms of known functions . The actual answer may be verified at -
http://integrals.wolfram.com/index.jsp?expr=%7BArcTan%5Bx%5D%7D%5E2&random=false
49@Soumya: This is an AISSCE question! :P
:P
This is then a sure boomerang! must be awarding free +6 to all??
39This came in the boards some years back and if I remember correctly, there was controversy regarding it. Surely +6 to all.
1use the following word to solve by parts
ILATE
I - Inverse trignometric functions
L - Logarithmic functions
A - Arithmatics
T - Trignometric functions
E - Exponential functions
take the above functions values by u
71@anish.. Yeah that is okay, But it won't work here.. You can even try ∫x tanx dx
