The integral \int { (x^8-x^2) \dx}/{ (x^{12}+3 x^6+1) \tan ^{-1} (x^3+\frac{1}{x^3}) } is equal to : [27-Jan-2024 Shift 2]

Correct Answer is : loge(|tan−1(x3+1x3)|)1∕3+C

\log _e ({| \tan ^{-1} (x^3+\frac{1}{x^3}) |}) ^{1 \/ 3}+C

\log _e ({| \tan ^{-1} (x^3+\frac{1}{x^3}) |}) ^{1 \/ 2}+C

\log _e ({| \tan ^{-1} (x^3+\frac{1}{x^3}) |}) +C

\log _e ({| \tan ^{-1} (x^3+\frac{1}{x^3}) |}) ^3+C

JEE Main 27-Jan-2024 Shift 2 Solved Paper

Section: Mathematics

Question : 14 of 90

Marks: +1, -0

The integral ∫‌

(x8−x2)‌dx

(x12+3x6+1)tan−1(x3+‌

)

is equal to :

[27-Jan-2024 Shift 2]

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