InterviewSolution
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InterviewSolution
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The value of `int_0^oox/((1+x)(x^2+1))dx` isA. ` 2pi `B. ` ( pi ) /( 4 ) `C. ` ( pi ) /( 16 ) `D. ` ( pi ) /( 32 ) ` |
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Answer» 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 ) ` |
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