JEE Mains · Maths · STD 12 - 7.1 indefinite integral
The integral \(\int \frac{\left(x^8-x^2\right) d x}{\left(x^{12}+3 x^6+1\right) \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)}\) is equal to :
- A \(\log _e\left(\left|\tan ^{-1}\left(\mathrm{x}^3+\frac{1}{\mathrm{x}^3}\right)\right|\right)^{1 / 3}+\mathrm{C}\)
- B \(\log _e\left(\left|\tan ^{-1}\left(x^3+\frac{1}{x^3}\right)\right|\right)^{1 / 2}+C\)
- C \(\log _e\left(\left|\tan ^{-1}\left(x^j+\frac{1}{x^j}\right)\right|\right)+C\)
- D \(\log _e\left(\left|\tan ^{-1}\left(x^3+\frac{1}{x^3}\right)\right|\right)^3+C\)
Answer & Solution
Correct Answer
(A) \(\log _e\left(\left|\tan ^{-1}\left(\mathrm{x}^3+\frac{1}{\mathrm{x}^3}\right)\right|\right)^{1 / 3}+\mathrm{C}\)
Step-by-step Solution
Detailed explanation
\( I=\int \frac{x^8-x^2}{\left(x^{12}+3 x^6+1\right) \tan ^{-1}\left(x^3+\frac{1}{x^3}\right.} \) \( \text { Let } \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)=t \) \( \Rightarrow \frac{1}{1+\left(x^3+\frac{1}{x^3}\right)^2} \cdot\left(3 x^2-\frac{3}{x^4}\right) d x \)…
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