MHT CET · Maths · Indefinite Integration
\(\int \sqrt{\frac{1+x}{1-x}} d x=\)
(where \(C\) is a constant of integration.)
- A \(\sin ^{-1} x-\sqrt{1-x^2}+C\)
- B \(\sqrt{1-x^2}-\sqrt{x}+C\)
- C \(-\sqrt{1-x^2}+\sqrt{1+x}+C\)
- D \(\sin ^{-1} x+\sqrt{1-x^2}+C\)
Answer & Solution
Correct Answer
(A) \(\sin ^{-1} x-\sqrt{1-x^2}+C\)
Step-by-step Solution
Detailed explanation
\(\int \sqrt{\frac{1+x}{1-x}} \mathrm{~d} x=\int \sqrt{\frac{(1+x)^2}{1-x^2}} \mathrm{~d} x=\int\{\frac{1}{\sqrt{1-x^2}}+\) \(\frac{x}{\sqrt{1-x^2}}\} \mathrm{d} x\)
\(=\sin ^{-1} x-\frac{1}{2} \times 2 \sqrt{1-x^2}\)
\(=\sin ^{-1} x-\sqrt{1-x^2}+C\)
\(=\sin ^{-1} x-\frac{1}{2} \times 2 \sqrt{1-x^2}\)
\(=\sin ^{-1} x-\sqrt{1-x^2}+C\)
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