JEE Main20204 Sep 2020Morning ShiftMathematicsIndefinite IntegrationActual
Let f x = ∫ x 1 + x 2 d x x ≥ 0 . Then f 3 - f 1 is equal to :
Options
- A- π 12 + 1 2 + 3 4
- Bπ 6 + 1 2 - 3 4
- C- π 6 + 1 2 + 3 4
- Dπ 12 + 1 2 - 3 4
Correct answer
D. π 12 + 1 2 - 3 4
Step-by-step solution
f x = ∫ x 1 + x 2 d x Let x = tan 2 θ d x = 2 tan θ sec 2 θ   d θ f x = ∫ tan θ 1 + tan 2 θ 2 . 2 tan θ sec 2 θ   d θ f x = ∫ tan θ sec 4 θ . 2 tan θ sec 2 θ   d θ f x = ∫ 2 tan 2 θ . cos 2 θ   d θ f x = ∫ 2 sin 2 θ   d θ f x = ∫ 1 − cos 2 θ   d θ f x = θ − sin 2 θ 2 + C = θ − tan θ 1 + tan 2 θ + C f x =