TS EAMCET · Maths · Indefinite Integration
\(\int \frac{x^2+1}{x^4+7 x^2+1} d x\) is equal to
- A \(\frac{1}{3} \tan ^{-1}\left(\frac{x^2-1}{3 x}\right)+C\)
- B \(\tan ^{-1}\left(\frac{x^2-1}{x}\right)+C\)
- C \(\frac{1}{3} \tan ^{-1}\left(\frac{x^2-1}{x}\right)+C\)
- D \(\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{x^2-1}{\sqrt{3} x}\right)+C\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{3} \tan ^{-1}\left(\frac{x^2-1}{3 x}\right)+C\)
Step-by-step Solution
Detailed explanation
Let \[ \begin{aligned} I & =\int \frac{x^2+1}{x^4+7 x^2+1} d x \\ & =\int \frac{1+\frac{1}{x^2}}{x^2+7+\frac{1}{x^2}} d x=\int \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}+7} d x \end{aligned} \]…
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