Q112 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics
If $\int \frac{1}{x^4+8 x^2+9} d x$ $=\frac{1}{k}\left[\frac{1}{\sqrt{14}} \tan ^{-1}(f(x))-\frac{1}{\sqrt{2}} \tan ^{-1}(g(x))\right]+c$,
then $\sqrt{\frac{k}{2}+f(\sqrt{3})+g(1)}=$
- A$3-2 \sqrt{2}$
- B$\sqrt{2}-1$
- C$\sqrt{3}+2 \sqrt{2}$
- Correct answerD$\sqrt{2}+1$
Solution
$\begin{aligned} & \mathrm{I}=\int \frac{1}{x^4+8 x^2+9} d x=\int \frac{\frac{1}{x^2}}{x^2+8+\frac{9}{x^2}} d x \\ & =\frac{1}{6} \int \frac{\left(1+\frac{3}{x^2}\right)}{\left(x-\frac{3}{x}\right)^2+14} d x-\frac{1}{6} \int \frac{\left(1-\frac{3}{x^2}\right)}{\left(x+\frac{3}{x}\right)^2+2} d x \\ & \text{ Let }x-\frac{3}{x}=t \text { and } x+\frac{3}{x}=\frac{1}{6} \int \frac{d t}{t^2+(\sqrt{14})^2}-\frac{1}{6} \int \frac{d(\mu}{\mu^2+(\sqrt{2})^2} \\ & \mathrm{I}=\frac{1}{6}\left[\frac{1}{\sqrt{14}} \tan ^{-1} \frac{t}{\sqrt{14}}-\frac{1}{\sqrt{2}} \tan ^{-1} \frac{\mu}{\sqrt{2}}\right]+c \\ & \Rightarrow \mathrm{k}=6, \mathrm{f}(\mathrm{x})=\frac{1}{\sqrt{14}}\left(x-\frac{3}{x}\right), g(x)=\frac{1}{\sqrt{2}}\left(x+\frac{3}{x}\right)\end{aligned}$
$\therefore \sqrt{\frac{k}{2}+f(\sqrt{3})+g(1)}=\sqrt{3+2 \sqrt{2}}=(\sqrt{1+\sqrt{2}})^2=\sqrt{2}+1$
Hence, the correct answer is option: 'D'.