AP EAMCET · Maths · Indefinite Integration
\(\int \frac{x^4+1}{x^6+1} d x=\)
- A \(\tan ^1 x-\tan ^{-1} x^3+c\)
- B \(\tan ^{-1} x-\frac{1}{3} \tan ^{-1} x^3+c\)
- C \(\tan ^{-1} x+\tan ^{-1} x^3+c\)
- 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
\(\begin{aligned} & \text { } \frac{x^4+1}{x^6+1}=\frac{\left(x^4-x^2+1+x^2\right)}{\left(x^6+1\right)}=\frac{1}{x^2+1}+\frac{x^2}{x^6+1} \\ & \int \frac{x^4+1}{x^6+1} d x=\int \frac{d x}{1+x^2}+\int \frac{x^2 d x}{x^6+1}=\tan ^{-1} x+\frac{1}{3} \tan ^{-1} x^3+c\end{aligned}\)
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