Question:
If \(\int \frac{dx}{(x^2 - 2x + 10)^2} = A\left(\tan^{-1}\left(\frac{x-1}{3}\right) + \frac{f(x)}{x^2 - 2x + 10}\right) + C\)
where \(C\) is a constant of integration then :
If \(\int \frac{dx}{(x^2 - 2x + 10)^2} = A\left(\tan^{-1}\left(\frac{x-1}{3}\right) + \frac{f(x)}{x^2 - 2x + 10}\right) + C\)
where \(C\) is a constant of integration then :
where \(C\) is a constant of integration then :
Show Hint
Complete the square, put \(x - 1 = 3\tan\theta\) and reduce to \(\int \cos^2\theta\, d\theta\).
Updated On: Nov 5, 2026
- \(A = \frac{1}{54}\) and \(f(x) = 9(x-1)^2\)
- \(A = \frac{1}{54}\) and \(f(x) = 3(x-1)\)
- \(A = \frac{1}{81}\) and \(f(x) = 3(x-1)\)
- \(A = \frac{1}{27}\) and \(f(x) = 9(x-1)^2\)
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The Correct Option is B
Solution and Explanation
\(x^2 - 2x + 10 = (x-1)^2 + 9\). Put \(x - 1 = 3\tan\theta\), \(dx = 3\sec^2\theta\, d\theta\).
\(I = \int \frac{3\sec^2\theta\, d\theta}{81\sec^4\theta} = \frac{1}{27}\int \cos^2\theta\, d\theta = \frac{1}{54}(\theta + \sin\theta\cos\theta) + C\).
Here \(\theta = \tan^{-1}\frac{x-1}{3}\), \(\sin\theta\cos\theta = \frac{\tan\theta}{1 + \tan^2\theta} = \frac{(x-1)/3}{1 + (x-1)^2/9} = \frac{3(x-1)}{x^2 - 2x + 10}\).
So \(I = \frac{1}{54}\left(\tan^{-1}\frac{x-1}{3} + \frac{3(x-1)}{x^2 - 2x + 10}\right) + C\), giving \(A = \frac{1}{54}\), \(f(x) = 3(x-1)\).
Answer: (2) \(A = \frac{1}{54}\) and \(f(x) = 3(x-1)\)
\(I = \int \frac{3\sec^2\theta\, d\theta}{81\sec^4\theta} = \frac{1}{27}\int \cos^2\theta\, d\theta = \frac{1}{54}(\theta + \sin\theta\cos\theta) + C\).
Here \(\theta = \tan^{-1}\frac{x-1}{3}\), \(\sin\theta\cos\theta = \frac{\tan\theta}{1 + \tan^2\theta} = \frac{(x-1)/3}{1 + (x-1)^2/9} = \frac{3(x-1)}{x^2 - 2x + 10}\).
So \(I = \frac{1}{54}\left(\tan^{-1}\frac{x-1}{3} + \frac{3(x-1)}{x^2 - 2x + 10}\right) + C\), giving \(A = \frac{1}{54}\), \(f(x) = 3(x-1)\).
Answer: (2) \(A = \frac{1}{54}\) and \(f(x) = 3(x-1)\)
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