Question

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

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

Hint:

Substitute t = tan inverse of x to the fourth power; then dt contains x cubed over 1 + x to the eighth.

Answer 1

The Correct Option is D

Step 1: Differentiate Option (D)
$\dfrac{d}{dx}\left[-\dfrac{1}{4}\cos(\tan^{-1}x^{4})\right]=\dfrac{1}{4}\sin(\tan^{-1}x^{4})\cdot\dfrac{4x^{3}}{1+x^{8}}$.

Step 2: Simplify
This equals $\dfrac{x^{3}\sin(\tan^{-1}x^{4})}{1+x^{8}}$, the given integrand.

Step 3: Compare
Options (A), (B) and (C) have the wrong sign or function, so (D) is correct.

Final Answer:
The correct option is (D). \[ \boxed{\text{(D) }-\dfrac{1}{4}\cos (\tan ^{-1}(x^{4}))+c} \]