NTA Abhyas JEE Main2020MathematicsDefinite IntegrationPractice
The value of the integral ∫ 0 1 4 t 3 1 + t 8 + 8 t 4 1 + t 7 d t is
Options
- A128
- B512
- C256
- D1024
Correct answer
C. 256
Step-by-step solution
Note that d d t t 4 1 + t 8 = 4 t 3 1 + t 8 + 8 t 4 1 + t 7 Hence, the required integral is t 4 1 + t 8 0 1 = 2 8