Prove the following

Question:

If $\int \frac{d x}{\left(x^{2}-2 x+10\right)^{2}}$

$=\mathrm{A}\left(\tan ^{-1}\left(\frac{\mathrm{x}-1}{3}\right)+\frac{f(\mathrm{x})}{\mathrm{x}^{2}-2 \mathrm{x}+10}\right)+\mathrm{C}$

where $\mathrm{C}$ is a constant of integration, then :

  1. $\mathrm{A}=\frac{1}{27}$ and $f(\mathrm{x})=9(\mathrm{x}-1)$

  2. $\mathrm{A}=\frac{1}{81}$ and $f(\mathrm{x})=3(\mathrm{x}-1)$

  3. $\mathrm{A}=\frac{1}{54}$ and $f(\mathrm{x})=9(\mathrm{x}-1)^{2}$

  4. $\mathrm{A}=\frac{1}{54}$ and $f(\mathrm{x})=3(\mathrm{x}-1)$


Correct Option: , 4

Solution:

$\int \frac{\mathrm{dx}}{\left((\mathrm{x}-1)^{2}+9\right)^{2}}=\frac{1}{27} \int \cos ^{2} \theta \mathrm{d} \theta \quad$ (Put $\quad \mathrm{x}-1=$

$3 \tan \theta)$

$=\frac{1}{54} \int(1+\cos 2 \theta) d \theta=\frac{1}{54}\left(\theta+\frac{\sin 2 \theta}{2}\right)+C$

$=\frac{1}{54}\left(\tan ^{-1}\left(\frac{x-1}{3}\right)+\frac{3(x-1)}{x^{2}-2 x+10}\right)+C$

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