∵7−6x−x2=16−(x+3)2 and dxd(7−6x−x2)=−2x−6 So, ∫7−6x−x22x+5dx=∫7−6x−x22x+6dx −∫16−(x+3)21dx =−27−6x−x2−sin−1(4x+3)+C Therefore, A=−2,&B=−1.
∫ 2 x+5 √7-6 x-x^2 d x=A √7-6 x-x^2+B sin ^-1 ( x+3 4 )+C (where C is a constant of integration), then the ordered pair (A, B) is equal to
JEE Main 2018 — Mathematics Calculus
∫7−6x−x22x+5dx=A7−6x−x2+Bsin−1(4x+3)+C (where C is a constant of integration), then the ordered pair (A,B) is equal to
Held on 15 Apr 2018 · Verified 6 Jul 2026.
Options
- A
(−2,−1)
- B
(2,−1)
- C
(−2,1)
- D
(2,1)
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