Question

If I = $\sqrt{\frac{1 + x}{1 + x}}$ dx, then I equals:

• A. log |x| + log $\left| 1 + \sqrt{1 - x^{2}} \right|$+ sin^–1$\sqrt{x}$+ C
• B. log |x| – log $\left| 1 - \sqrt{1 - x^{2}} \right|$+ tan^–1 x + C
• C. log |x| – log $\left| 1 + \sqrt{1 - x^{2}} \right|$– sin^–1 x + C
• D. log$\left| 1 + \sqrt{1 - x^{2}} \right|$– log |x| + cos^–1 x + C

Answer 1

Answer: (C)

log |x| – log $\left| 1 + \sqrt{1 - x^{2}} \right|$– sin^–1 x + C

Answer 2

I = $\frac{1 - x}{\sqrt{1 - x^{2}}}$dx

= $\int \frac { d x } { x \sqrt { 1 - x ^ { 2 } } }$ – $\int_{}^{}\frac{dx}{\sqrt{1 - x^{2}}}$

In the first integral, put x = $\frac{1}{t}$, so that

I = $\int \frac { \left( - 1 / \mathrm { t } ^ { 2 } \right) \mathrm { dt } } { \frac { 1 } { \mathrm { t } } \sqrt { 1 - \frac { 1 } { \mathrm { t } ^ { 2 } } } }$– sin^–1 x = – $\int_{}^{}\frac{dt}{\sqrt{t^{2} - 1}}$– sin^–1 x

= – log | t + $\sqrt{t^{2} - 1}$| – sin^–1 x + C

= – log $\left| \frac{1}{x} + \frac{\sqrt{1 - x^{2}}}{x} \right|$– sin^–1 x + C

= log |x| – log |1 + $\sqrt{1 - x^{2}}$| – sin^–1 x + C