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If ∫(x tan -1 x dx)/((1+x 2 ) 3/2 ) =(k 1 tan -1 x – k 2 )/(1+x 2 ) 1/2 +c where k 1 , k 2 are polynomial function of variable x and c is any real constant then value of 2 0 ∫max{k 1 ,k 2 }dx is (A)-2/3 (B)2/3 (C)-3/2 (D)-1/2

If ∫(x tan-1x dx)/((1+x2)3/2) =(k1tan-1x – k2)/(1+x2)1/2  +c where k1, k2 are polynomial function of variable x  and c is any real constant then value of 20∫max{k1,k2}dx is
(A)-2/3   (B)2/3    (C)-3/2   (D)-1/2

Grade:12

1 Answers

Vikas TU
14149 Points
6 years ago
Dear Student,
As Given in the qstn.=>
∫(x tan -1 x dx)/((1+x 2 ) 3/2 )= (x – tan-1(x) )/(1+x^2)^(1/2)
k1 = -1 , k2= -x
2 0 ∫max{k 1 ,k 2 }dx = -(1 0 ∫xdx + 1 2 ∫dx) = -(½ + 1) = -3/2 
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)

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