MHT CET · Maths · Indefinite Integration
\(\int \frac{x}{1+x^4} \mathrm{~d} x=\)
- A \(\frac{1}{2} \tan ^{-1} x^2+c\), where \(c\) is the constant of integration
- B \(2 \tan ^{-1} x+c\), where \(c\) is the constant of integration
- C \(\frac{1}{2} \tan ^{-1} x+c, \quad\) where \(c\) is the constant of integration
- D \(\tan ^{-1} x^2+c, \quad\) where \(c\) is the constant of integration
Answer & Solution
Correct Answer
(A) \(\frac{1}{2} \tan ^{-1} x^2+c\), where \(c\) is the constant of integration
Step-by-step Solution
Detailed explanation
Let \(u=x^2 \Rightarrow \mathrm{d}u = 2x \, \mathrm{d}x\) \(\int \frac{x}{1+x^4} \, \mathrm{d}x = \frac{1}{2} \int \frac{1}{1+u^2} \, \mathrm{d}u\)
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