Question

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\left(x\right)=9\left(x-1\right) $
• B. $A = \frac{1}{81} $ and $ f\left(x\right)= 3\left(x-1\right) $
• C. $A = \frac{1}{54} $ and $ f\left(x\right)=9\left(x-1\right)^2 $
• D. $A = \frac{1}{54} $ and $ f\left(x\right)= 3 \left(x-1\right) $

Answer 1

Answer: (D)

$\int \frac{dx}{\left(\left(x-1\right)^{2} +9\right)^{2}} = \frac{1}{27} \int\cos^{2} \theta d\theta \left(\text{Put} \quad x-1 = 3\tan\theta\right) $
$ = \frac{1}{54} \int \left(1+\cos2\theta\right) d\theta = \frac{1}{54} \left( \theta + \frac{\sin2\theta}{2}\right)+C $
$ = \frac{1}{54} \left(\tan^{-1} \left(\frac{x-1}{3}\right) + \frac{3\left(x-1\right)}{x^{2}-2x+10}\right) +C $