AP EAMCET · Maths · Definite Integration
\(\int_1^2 \frac{x^4-1}{x^6-1} d x=\)
- A \(\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)
- B \(\frac{121}{6}\)
- C \(\sqrt{2}-1\)
- D \(\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{2}{\sqrt{3}}\right)\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)
Step-by-step Solution
Detailed explanation
Let \(I=\int_1^2 \frac{x^4-1}{x^6-1} d x\) \(\begin{aligned} & =\int_1^2 \frac{(x-1)(x+1)\left(x^2+1\right)}{(x-1)\left(x^2+x+1\right)(x+1)\left(x^2-x+1\right)} d x \\ & =\int_1^2 \frac{x^2+1}{\left(x^2+x+1\right)\left(x^2-x+1\right)} d x\end{aligned}\)…
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