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JEE Mains · Maths · STD 12 - 7.1 indefinite integral
\(\int {\frac{{2x + 5}}{{\sqrt {7 - 6x - {x^2}} }}dx} = A\sqrt {7 - 6x - {x^2}} + B\,{\sin ^{ - 1}}\left( {\frac{{x + 3}}{4}} \right) + C\) (where \(C\) is a constant of integration), then the ordered pair \((A, B)\) is equal to
- A \((- 2, - 1 )\)
- B \((2, - 1 )\)
- C \((- 2, 1)\)
- D \((2, 1)\)
Answer & Solution
Correct Answer
(A) \((- 2, - 1 )\)
Step-by-step Solution
Detailed explanation
\(\int \frac{2 x+5}{\sqrt{7-6 x-x^{2}}} \cdot d x\) \(\text { Let } 7-6 x-x^{2}=t^{2}\) \((-6,-2 x) \cdot d x=2 t . d t\) \(-\int \frac{2 t . d t}{t}-\int \frac{1}{\sqrt{16-(x+3)^{2}}} d x\) \(-2 t-\sin ^{-1}\left(\frac{x+3}{4}\right)+C\) \(A=-2, B=-1\)
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