TS EAMCET · Maths · Indefinite Integration
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 \(\sqrt{3}+2 \sqrt{2}\)
- D \(\sqrt{2}+1\)
Answer & Solution
Correct Answer
(D) \(\sqrt{2}+1\)
Step-by-step Solution
Detailed explanation
\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…
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