Question
Let f(x)= x^2 d x (1+x^2 ) (1+ .1+x^2 ) . and f(0)=0, then the value of f(1) be
BITSAT
Mathematics
Indefinite Integration
2022
BITSAT 2022
BITSAT Mathematics Question (2022) — Solution
Options
- A. (1+ 2 )
- B. (1+ 2 )- 4
- C. (1+ 2 )+ 2
- D. None of these
Answer
B. (1+ 2 )- 4
Step-by-step solution
f(x)= x^2 d x (1+x^2 ) (1+ 1+x^2 ) Let x= d x= ^2 d = (1+x^2 ) d f(x)= x^2 d x (1+x^2 ) (1+ 1+x^2 ) = ^2 ^2 d ^2 (1+ ) = ^2 d 1+ = ^2 d (1+ ) = 1- ^2 d (1+ ) = (1- ) d = d - d = (x+ 1+x^2 )- ^ -1 x+C f(0)= (0+ 1+0 )- ^ -1 (0)+C 0= 1-0+C C=0 f(1)= (1+ 1+1^2 )- ^ -1 (1) = (1+ 2 )- 4
Related: Mathematics — Indefinite Integration · All PYQ Banks