JEE Main2005MathematicsIndefinite IntegrationActual

( x-1) (1+( x)^2 . ^2 d x is equal to

Options

Ax ( x)^2+1 +C
Bx x^2+1 +C
Cx e^x 1+x^2 +C
Dx ( x)^2+1 +C

Correct answer

D. x ( x)^2+1 +C

Step-by-step solution

aligned & ( x-1)^2 (1+( x)^2 )^2 d x & = [ 1 (1+( x)^2 ) - 2 x (1+( x)^2 )^2 ] d x & = [ e^t 1+t^2 - 2 t e^t (1+t^2 )^2 ] d t put x=t d x=e^t d t & e^t [ 1 1+t^2 - 2 t (1+t^2 )^2 ] d t & = e^t 1+t^2 +c= x 1+( x)^2 +c aligned

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