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

Indefinite Integrals

Section: Mathematics

Question : 22 of 89

Marks: +1, -0

The integral ∫‌

(x8−x2)‌dx

(x12+3x6+1)tan−1(x3+‌

)

is equal to :

[27-Jan-2024 Shift 2]

Go to Question:

Prev Question

Next Question