JEE Main202427 Jan 2024Evening ShiftMathematicsIndefinite IntegrationActual
The integral ∫ x 8 - x 2 dx x 12 + 3 x 6 + 1 tan - 1 x 3 + 1 x 3 is equal to :
Options
- Alog tan - 1 x 3 + 1 x 3 1 3 + C
- Blog e tan - 1 x 3 + 1 x 3 1 2 + C
- Clog e tan - 1 x 3 + 1 x 3 + C
- Dlog e tan - 1 x 3 + 1 x 3 3 + C
Correct answer
A. log tan - 1 x 3 + 1 x 3 1 3 + C
Step-by-step solution
Let, I = ∫ x 8 - x 2 x 12 + 3 x 6 + 1 tan - 1 x 3 + 1 x 3 dx Putting, tan - 1 x 3 + 1 x 3 = t ⇒ 1 1 + x 3 + 1 x 3 2 · 3 x 2 - 3 x 4 dx = dt ⇒ 1 1 + x 6 + 1 x 6 + 2 · 3 x 2 - 3 x 4 dx = dt ⇒ x 6 x 12 + 3 x 6 + 1 · 3 x 6 - 3 x 4 dx = dt ⇒ x 8 - x 2 x 12 + 3 x 6 + 1 dx = dt 3 ⇒ I = 1 3 ∫ 1 t dt ⇒ I = 1 3 log | t | + C ⇒ I = 1 3 log tan - 1 x 3 + 1 x 3 + C ⇒ I = log tan - 1 x 3 + 1 x 3 1 3 + C Hence optin A is correct