\int \frac{x^3 \tan ^{-1} x^4}{1+x^8} \dx= [AP EAPCET 22 May 2024 Shift 1]

Correct Answer is : (Tan−1(x4))28+c

{ ({\Tan}^{-1} (x^4) ) ^2}/{8}+c

{ ({\Tan}^{-1} (x^4) ) ^3}/{3}+c

{ ({\Tan}^{-1} (x^4) ) ^2}/{4}+c

{ ({\Tan}^{-1} (x^4) ) ^2}/{2}+c

Indefinite Integration

Section: Mathematics

Question : 55 of 89

Marks: +1, -0

∫‌

x3tan−1x4

1+x8

‌dx=

[AP EAPCET 22 May 2024 Shift 1]

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