Question

\(\int\frac{x^3\sin(\tan^{-1}(x^4))}{1+x^8}dx\) is equal to

• A. \(\frac{-\cos(\tan^{-1}(x^4))}{4}+C\)
• B. \(\frac{\cos(\tan^{-1}(x^4))}{4}+C\)
• C. \(\frac{-\cos(\tan^{-1}(x^3))}{3}+C\)
• D. \(\frac{-\sin(\tan^{-1}(x^4))}{4}+C\)

Answer 1

The Correct Option is A

The correct answer is (A) : \(\frac{-\cos(\tan^{-1}(x^4))}{4}+C\).