99 Percentile Qs Bank for JEE MainMathematicsIndefinite Integration
If x x - [3] x d x=x+E x^ 5 / 6 +D x^ 2 / 3 +C x^ 1 / 2 +B x^ 1 / 3 +A x^ 1 / 6 + ( [6] x -1)^6+K , then A+B+C+D+E=
Options
- A137 10
- B129 10
- C119 10
- D117 10
Correct answer
A. 137 10
Step-by-step solution
Let I= x x - [3] x d x Put x=t^6 d x=6 t^5 d t aligned & I= t^3 6 t^5 t^3-t^2 d t & =6 t^8 t^2(t-1) d t=6 t^6 t-1 d t & =6 (t^6-1 )+1 t-1 d t=6 (t^3-1 ) (t^3+1 )+1 t-1 d t & =6 (t-1) (t^2+t+1 ) (t^3+1 )+1 t-1 d t & =6 [t^5+t^2+t^4+t+t^3+1+ 1 t-1 ] d t & =6 [t^5+t^4+t^3+t^2+t+1+ 1 t-1 ] d t & =6 [ t^6 6 + t^5 5 + t^4 4 + t^3 3 + t^2 2 +t+ (t-1) ]+k & =x+ 6 5 x^ 5 / 6 + 3 2 x^ 2 / 3 +2 x^ 3 / 2 +3 x^ 3 / 3 & +6 x^ 1 / 6 + Iog ( [6] x-1 )^6+k & aligned aligned & A=6, B=3, C=2 D= 3 2 B= 6 5 & A+B+C+D+B=6+3+2+ 3 2 + 6 5