WBJEE · Maths · Indefinite Integration
\(\int \frac{x^3 d x}{1+x^8}=\)
- A \(4 \tan ^{-1} \mathrm{x}^3+\mathrm{c}\)
- B \(\frac{1}{4} \tan ^{-1} x^4+c\)
- C \(x+4 \tan ^{-1} x^4+c\)
- D \(x^2+\frac{1}{4} \tan ^{-1} x^4+c\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{4} \tan ^{-1} x^4+c\)
Step-by-step Solution
Detailed explanation
Hints: \(\int \frac{x^3 d y}{1+\left(x^4\right)^2}=\frac{1}{4} \tan ^{-1}\left(x^4\right)\)
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