ExamBro
ExamBro
MHT CET · Maths · Indefinite Integration

\(\int \sqrt{\frac{1+x}{1-x}} d x=\)
(where \(C\) is a constant of integration.)

  1. A \(\sin ^{-1} x-\sqrt{1-x^2}+C\)
  2. B \(\sqrt{1-x^2}-\sqrt{x}+C\)
  3. C \(-\sqrt{1-x^2}+\sqrt{1+x}+C\)
  4. D \(\sin ^{-1} x+\sqrt{1-x^2}+C\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\sin ^{-1} x-\sqrt{1-x^2}+C\)

Step-by-step Solution

Detailed explanation

\(\int \sqrt{\frac{1+x}{1-x}} \mathrm{~d} x=\int \sqrt{\frac{(1+x)^2}{1-x^2}} \mathrm{~d} x=\int\{\frac{1}{\sqrt{1-x^2}}+\) \(\frac{x}{\sqrt{1-x^2}}\} \mathrm{d} x\)
\(=\sin ^{-1} x-\frac{1}{2} \times 2 \sqrt{1-x^2}\)
\(=\sin ^{-1} x-\sqrt{1-x^2}+C\)