AP EAMCET202018 Sep 2020Morning ShiftMathematicsIndefinite IntegrationActual
( x-1 (x x+1 )⁻¹ d x= )
Options
- A( |x+ x^2-1 |- ⁻¹(x)+c )
- B( |x- x^2-1 |- ⁻¹(x)+c )
- C( |x+ x^2-1 |+ ⁻¹(x)+c )
- D( |x+ x^2-1 |- ⁻¹(x)+c )
Correct answer
A. ( |x+ x^2-1 |- ⁻¹(x)+c )
Step-by-step solution
(I= x-1 (x x+1 )⁻¹ d x= 1 x x-1 x+1 d x ) Put ( x-1 x+1 =t^2 x= 1+t^2 1-t^2 ) So, (d x= (1-t^2 )(2 t)- (l+t^2 )(-2 t) (1-t^2 )^2 d t ) ( d x= 4 t (1-t^2 )^2 d t ) So, ( I= ( 1-t^2 1+t^2 ) t 4 t (1-t^2 )^2 d t= 4 t^2 (1+t^2 ) (1-t^2 ) d t ) ( aligned & =2 ( 1 1-t^2 - 1 .1+t^2 ) ) d t & =2 [- 1 2 ( 1-t 1+t )- ⁻¹(t) ]+c aligned ) (=- _e ( 1- x-1 x+1 1+ x-1 x+1 )-2 ⁻¹ x-1 x+1 +c ) (=- _e ( x+1 - x-1 x+1 + x-1 )- ⁻¹ ( 2 x-1 x+1 1- x-1 x+1 )+c ) (=- _e ( x+1+x-1-2 x^2-1 (x+1)-(x-1) )- ⁻¹ x^2-1 +c ) (=- _e (x- x^2-1 )- ⁻¹