indefinite-integration

Question: $\int{\sqrt{\frac{1+x}{1-x}}\,\,dx=}$

1) $-{{\sin }^{-1}}x-\sqrt{1-{{x}^{2}}}\,+c$

2) ${{\sin }^{-1}}x+\sqrt{1-{{x}^{2}}}\,+c$

3) ${{\sin }^{-1}}x-\sqrt{1-{{x}^{2}}}\,+c$

4) $-{{\sin }^{-1}}x-\sqrt{{{x}^{2}}-1}\,+c$

Solution:

Explanation: No Explanation

Integration by substitution

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