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

Question

$-\frac{\cos(\tan^{-1}(x^4))}{4} + C$

$\frac{\cos(\tan^{-1}(x^4))}{4} + C$

$-\frac{\cos(\tan^{-1}(x^3))}{3} + C$

$\frac{\sin(\tan^{-1}(x^4))}{4} + C$

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