AP EAMCET · Maths · Indefinite Integration
\(\int \frac{x^4+1}{1+x^6} d x=\)
- A \(\tan ^{-1}\left(x^3\right)+\tan ^{-1} x+c\)
- B \(\frac{1}{3} \tan ^{-1} x+\tan ^{-1} x^3+c\)
- C \(3 \tan ^{-1} x^3+\tan ^{-1} x+c\)
- D \(\tan ^{-1} x+\frac{1}{3} \tan ^{-1} x^3+c\)
Answer & Solution
Correct Answer
(D) \(\tan ^{-1} x+\frac{1}{3} \tan ^{-1} x^3+c\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \int \frac{x^4+1}{1+x^6} d x=\int \frac{x^4-x^2+1+x^2}{1+x^6} d x \\ & =\int \frac{x^4-x^2+1}{1+x^6} d x+\int \frac{x^2}{1+x^6} d x \\ & I_1 \quad I_2 \\ & I_1=\int \frac{x^4-x^2+1}{1+\left(x^2\right)^3} d x=\int…
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