Correct Answer - B
Let ` I = int _ 0 ^( oo) ( x dx ) /( (1 + x ) ( x^ 2 + 1 ) ) `
by partial fraction,
` ( x )/( ( 1 + x ) ( x^ 2 + 1 ) ) = ( A) / ( ( 1 + x ) ) + (B x + C ) /( ( x ^ 2 + 1 )) `
` rArr x = A ( x^ 2 + 1 ) + ( 1 + x ) (Bx +C ) `
` rArr x= A ( x^ 2 + 1 ) + ( B x + Bx ^ 2 + C + Cx ) `
` rArr x = ( A +B ) x^ 2 + ( B + C ) x + ( A+C ) `
On comparing both sides , we get
` A + B = 0 , B +C = 1, A +C= 0 ` ... (i)
On adding all these equations , we get
` A + B+C = ( 1 ) / ( 2 ) " " `... (ii)
` therefore A = (1 ) / ( 2 ) - 1 = - ( 1 ) / ( 2 ) , C= ( 1 )/ ( 2 ) `
and ` B = ( 1 ) / (2 ) `
` therefore I = int _ 0 ^( oo) { ( - 1 ) /( 2 ( 1 + x ) ) + ( 1 ) / ( 2) ( ( x + 1 ))/ ( ( x^ 2 + 1 ))} dx `
` = - ( 1 ) / ( 2 ) int _ 0 ^( oo) ( dx ) / ( 1 + x ) + (1 ) / ( 2 ) int _ 0 ^( oo) ( x ) /( x ^ 2 + 1 ) dx + ( 1 ) / (2 ) int _ 0^(oo) ( dx ) / ( 1 + x ^ 2 ) `
` =- (1 ) / ( 2 ) [log ( 1 + x ) ]_0 ^(oo) + (1 )/ ( 4 ) [log ( x ^ 2 + 1 ) ] _0 ^(oo) + ( 1 )/ (2 ) xx ( pi ) / (2 ) `
` = - ( 1 ) / ( 2 ) lim _ ( xto oo ) log ( 1 + x ) + (1 )/ ( 4 )`
` lim_ ( x to oo) log ( 1 + x ^2 ) + ( pi )/(4 ) `
` = lim _ ( xto oo) log [ ( ( 1 + x^ 2 ) ^ (1//4) ) / ( ( 1 + x ) ^ ( 1// 2 ) ) ] + ( pi ) /( 4 ) `
` = lim_ ( x to oo ) log [ (sqrt x (( 1 )/ ( x^ 2) + 1 ) ^ (1// 4) ) / ( sqrt x ( ( 1 )/ ( x ) + 1 )^ (1// 2 ) ) ] + (pi )/ ( 4 ) `
` =log ""( ( 0 + 1 ) ^ (1//4))/( ( 0 + 1 )^(1//2 ) ) + ( pi) / ( 4 ) `
` = log ( 1 ) + ( pi ) / (4 ) = 0 + (pi ) /(4 ) = ( pi ) / ( 4 ) `
