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JEE Mains · Maths · STD 12 - 7.2 definite integral
If \(\int\limits_1^2 {\frac{{dx}}{{{{\left( {{x^2} - 2x + 4} \right)}^{\frac{3}{2}}}}} = \frac{k}{{k + 5}}} \) then \(k\) is equal to
- A \(1\)
- B \(2\)
- C \(3\)
- D \(4\)
Answer & Solution
Correct Answer
(A) \(1\)
Step-by-step Solution
Detailed explanation
\(\int_{1}^{2} \frac{d x}{\left((x-1)^{3}+3^{3 / 2}\right.}\) \(x-1=\sqrt{3} \tan \theta\) \(d x=\sqrt{3} \sec ^{2} \cdot d \theta\) \(\int_{0}^{\pi / 6} \frac{\sqrt{3} \sec ^{2} \theta d \theta}{\left(\sqrt{3} \tan \theta^{2}+(\sqrt{3})^{2}\right)^{3 / 2}}\)…
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