TS EAMCET · Maths · Indefinite Integration
\(\int \frac{x^{49} \tan ^{-1}\left(x^{50}\right)}{\left(1+x^{100}\right)} d x=k\left(\tan ^{-1}\left(x^{50}\right)\right)^2+c\), then \(k\) is equal to
- A \(\frac{1}{50}\)
- B \(-\frac{1}{50}\)
- C \(\frac{1}{100}\)
- D \(-\frac{1}{100}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{100}\)
Step-by-step Solution
Detailed explanation
Let \(I=\int x^{49} \frac{\tan ^{-1}\left(x^{50}\right)}{1+\left(x^{50}\right)^2} d x\) Let \(x^{50}=t \Rightarrow 50 x^{49} d x=d t\) \(\therefore \quad I=\frac{1}{50} \int \frac{\tan ^{-1} t}{1+t^2} d t\) Let \(\tan ^{-1} t=u \Rightarrow \frac{1}{1+t^2} d t=d u\)…
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