JEE Advanced2008MathematicsIndefinite IntegrationActual
Let I= e^x e^ 4 x +e^ 2 x +1 d x, J= e^ -x e^ -4 x +e^ -2 x +1 d x . Then, for an arbitrary constant C , the value of J-I equals
Options
- A1 2 | e^ 4 x -e^ 2 x +1 e^ 4 x +e^ 2 x +1 |+C
- B1 2 | e^ 2 x +e^x+1 e^ 2 x -e^x+1 |+C
- C1 2 | e^ 2 x -e^x+1 e^ 2 x +e^x+1 |+C
- D1 2 | e^ 4 x +e^ 2 x +1 e^ 4 x -e^ 2 x +1 |+C
Correct answer
C. 1 2 | e^ 2 x -e^x+1 e^ 2 x +e^x+1 |+C
Step-by-step solution
Since, J= e^ 3 x 1+e^ 2 x +e^ 4 x d x aligned J-I & = (e^ 3 x -e^x ) 1+e^ 2 x +e^ 4 x d x= (u^2-1 ) 1+u^2+u^4 d u & = (1- 1 u^2 ) 1+ 1 u^2 +u^2 d u= (1- 1 u^2 ) (u+ 1 u )^2-1 d u & = d t t^2-1 & = 1 2 | t-1 t+1 |+C & = 1 2 | u^2-u+1 u^2+u+1 |+C= 1 2 | e^ 2 x -e^x+1 e^ 2 x +e^x+1 |+C aligned