JEE Mains · Maths · STD 12 - 7.1 indefinite integral
If \(\int {\frac{{dx}}{{{{\left( {{x^2} - 2x + 10} \right)}^2}}} = A\left( {{{\tan }^{ - 1}}\left( {\frac{{x - 1}}{3}} \right) + \frac{{f\left( x \right)}}{{{x^2} - 2x + 10}}} \right)} + C\) Where \(C\) is a constant of integration, then
- A \(A = \frac{1}{{27}}\) and \(f(x) = -(x - 1)\)
- B \(A = \frac{1}{{54}}\) and \(f(x) = 9(x - 1)^2\)
- C \(A = \frac{1}{{54}}\) and \(f(x) = 3(x - 1)\)
- D \(A = \frac{1}{{81}}\) and \(f(x) = 3(x - 1)\)
Answer & Solution
Correct Answer
(C) \(A = \frac{1}{{54}}\) and \(f(x) = 3(x - 1)\)
Step-by-step Solution
Detailed explanation
\(\int \frac{d x}{\left(x^{2}-2 x+10\right)^{2}}=\int \frac{d x}{\left((x-1)^{21}+9\right)^{2}}\) \(\text { Let } x-1=3 \tan \theta\) \({\mathrm{d} x=3 \sec ^{2} \theta \mathrm{d} \theta} \) \(\therefore \frac{1}{{27}}\int {{{\cos }^2}} \theta {\rm{d}}\theta \)…
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