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)

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