AP EAMCET201922 Apr 2019Morning ShiftMathematicsIndefinite IntegrationActual
( 2 x 4 x ^4 x (1+ ^2 2 x ) d x= )
Options
- A( ( 1+ 2 x 1+ ^2 2 x )+ ^2 x+c )
- B( (1+ 2 x)^2 (1+ ^2 x ) + x+c )
- C( (1+ 2 x)^2 (1+ ^2 2 x ) + ^2 x+c )
- D( 1+ ^2 2 x (1+ 2 x)^2 + x+c )
Correct answer
C. ( (1+ 2 x)^2 (1+ ^2 2 x ) + ^2 x+c )
Step-by-step solution
(I=4 2 x (2 2 x 2 x) (1+ 2 x)^2 (1+ ^2 2 x ) d x ) Put ( 2 x=t ) ( -2 2 x d x=d t ) ( I=-4 t^2 (1+t)^2 (1+t^2 ) d t ) Split into partial fractions ( aligned I & =-4 [ 1 2(1+t)^2 - 1 2(1+t) + t 2 (1+t^2 ) ] d t & =-2 [- 1 1+t - (1+t)+ 1 2 (1+t^2 ) ] & = 2 2 ^2 x +2 (1+ 2 x)- (1+ ^2 2 x )+c & = ^2 x+2 (1+ 2 x)- (1+ ^2 2 x )+c & = ^2 x+ (1+ 2 x)^2 (1+ ^2 2 x ) +c aligned )