AP EAMCET · Maths · Indefinite Integration
\(\int \operatorname{Cos}^{-1}\left(\frac{1-x^2}{1+x^2}\right) d x=\)
- A \(2\left[x \operatorname{Tan}^{-1} x-\log \sqrt{1+x^2}\right]+c\)
- B \(2x \operatorname{Tan}^{-1} x+\log \sqrt{1-x^2}+c\)
- C \(x \operatorname{Tan}^{-1} x+\log \sqrt{1-x^2}+c\)
- D \(2\left[\operatorname{Tan}^{-1} x-\log \sqrt{1+x^2}\right]+c\)
Answer & Solution
Correct Answer
(A) \(2\left[x \operatorname{Tan}^{-1} x-\log \sqrt{1+x^2}\right]+c\)
Step-by-step Solution
Detailed explanation
\(\operatorname{Cos}^{-1}\left(\frac{1-x^2}{1+x^2}\right) = 2 \operatorname{Tan}^{-1} x\) \(\int 2 \operatorname{Tan}^{-1} x \, dx = 2 \int 1 \cdot \operatorname{Tan}^{-1} x \, dx\) \(2\left[x \operatorname{Tan}^{-1} x - \int x \frac{1}{1+x^2} dx\right]\)…
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