AP EAMCET · Maths · Indefinite Integration
\(\int \sqrt{x-1}(x \sqrt{x+1})^{-1} d x=\)
- A \(\ln \left|x+\sqrt{x^2-1}\right|-\sec ^{-1}(x)+c\)
- B \(\ln \left|x-\sqrt{x^2-1}\right|-\tan ^{-1}(x)+c\)
- C \(\ln \left|x+\sqrt{x^2-1}\right|+\sec ^{-1}(x)+c\)
- D \(\ln \left|x+\sqrt{x^2-1}\right|-\tan ^{-1}(x)+c\)
Answer & Solution
Correct Answer
(A) \(\ln \left|x+\sqrt{x^2-1}\right|-\sec ^{-1}(x)+c\)
Step-by-step Solution
Detailed explanation
\(I=\int \sqrt{x-1}(x \sqrt{x+1})^{-1} d x=\int \frac{1}{x} \cdot \sqrt{\frac{x-1}{x+1}} d x\) Put \(\frac{x-1}{x+1}=t^2 \Rightarrow x=\frac{1+t^2}{1-t^2}\) So, \(d x=\frac{\left(1-t^2\right)(2 t)-\left(l+t^2\right)(-2 t)}{\left(1-t^2\right)^2} d t\)…
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