TS EAMCET · Maths · Indefinite Integration
\(\int \frac{x^2}{1+x^6} d x\) is equal to
- A \(x^3+C\)
- B \(\frac{1}{3} \tan ^{-1}\left(x^3\right)+C\)
- C \(\log \left(1+x^3\right)\)
- D \(\frac{1}{1+x^3}+C\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{3} \tan ^{-1}\left(x^3\right)+C\)
Step-by-step Solution
Detailed explanation
Let \(I=\int \frac{x^2}{1+\left(x^3\right)^2} d x\) Put \(x^3=z \Rightarrow 3 x^2 d x=d z\) \[ I=\frac{1}{3} \int \frac{d z}{1+z^2}=\frac{1}{3} \tan ^{-1}(z)+C=\frac{1}{3} \tan ^{-1}\left(x^3\right)+C \]
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