AP EAMCET · Maths · Indefinite Integration
\(\int \frac{d x}{x+\sqrt{x-1}}=\)
- A \(\log _e|x+\sqrt{x-1}|-\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \sqrt{x-1}+1}{\sqrt{3}}\right)+c\)
- B \(\frac{1}{\sqrt{3}} \log _e|x+\sqrt{x-1}|-\tan ^{-1}\left(\frac{2 \sqrt{x-1}+1}{\sqrt{3}}\right)+c\)
- C \(\frac{2}{\sqrt{3}} \log _6|x+\sqrt{x-1}|-\tan ^{-1}\left(\frac{2 \sqrt{x-1}+1}{\sqrt{3}}\right)+c\)
- D \(\log _6|x+\sqrt{x-1}|-\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \sqrt{x-1}+1}{\sqrt{3}}\right)+c\)
Answer & Solution
Correct Answer
(D) \(\log _6|x+\sqrt{x-1}|-\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \sqrt{x-1}+1}{\sqrt{3}}\right)+c\)
Step-by-step Solution
Detailed explanation
Given integral, \(I=\int \frac{d x}{x+\sqrt{x-1}}\) put \(x-1=t^2 \Rightarrow d x=2 t d t\) then \(I=\int \frac{2 t}{\left(t^2+1\right)+t} d t=\int \frac{(2 t+1)-1}{t^2+t+1} d t\)…
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