Question

If $\int \frac{dx}{(x^2-2x+10{)}^2}=A({\tan }^{-1}\left(\frac{x-1}{3}\right)+\frac{f\left(x\right)}{x^2-2x+10})+C$ , where $C$ is a constant of integeration, then :

• A. $A=\frac{1}{81}$ and $f\left(x\right)=3(x-1)$
• B. $A=\frac{1}{27}$ and $f\left(x\right)=9(x-1)$
• C. $A=\frac{1}{54}$ and $f\left(x\right)=3(x-1)$
• D. $A=\frac{1}{54}$ and $f\left(x\right)=9(x-1{)}^2$

Answer 1

The correct option is C $A=\frac{1}{54}$ and $f\left(x\right)=3(x-1)$

$I=\int \frac{dx}{(x^2-2x+10{)}^2}=\int \frac{dx}{\left(\right(x-1{)}^2+9{)}^2}$

Put $x-1=3\tan t\Rightarrow dx=3{\sec }^{2}tdt$

$\Rightarrow I=\int \frac{3{\sec }^{2}tdt}{(9{\tan }^{2}t+9{)}^2}=\int \frac{3{\sec }^{2}tdt}{81{\sec }^{4}t}=\int \frac{{\cos }^{2}t}{27}dt=\int \frac{1+\cos 2t}{54}dt=\frac{1}{54}(t+\frac{\sin 2t}{2})+C=\frac{1}{54}[{\tan }^{-1}\left(\frac{x-3}{3}\right)+\frac{3(x-1)}{x^2-2x+10}]+C$

$\therefore A=\frac{1}{54},f\left(x\right)=3(x-1)$