A) \[2\left[ \sqrt{{{e}^{x}}-1}-{{\tan }^{-1}}\sqrt{{{e}^{x}}-1} \right]+c\]
B) \[\sqrt{{{e}^{x}}-1}-{{\tan }^{-1}}\sqrt{{{e}^{x}}-1}+c\]
C) \[\sqrt{{{e}^{x}}-1}+{{\tan }^{-1}}\sqrt{{{e}^{x}}-1}+c\]
D) \[2\left[ \sqrt{{{e}^{x}}-1}+{{\tan }^{-1}}\sqrt{{{e}^{x}}-1} \right]+c\]
E) \[2\left[ \sqrt{{{e}^{x}}-1}-{{\tan }^{-1}}\sqrt{{{e}^{x}}+1} \right]+c\]
Correct Answer: A
Solution :
\[A=\int{\sqrt{{{e}^{x}}-1}}\,dx\] Let \[{{e}^{x}}-1={{t}^{2}}\] Þ \[{{e}^{x}}dx=2t\,dt\]. Hence \[dx=\frac{2t}{{{t}^{2}}+1}dt\] \ \[A=\int{t\frac{2t}{{{t}^{2}}+1}}dt=\int{\frac{2{{t}^{2}}}{{{t}^{2}}+1}dt}\] \[=\int{\frac{2({{t}^{2}}+1)-2}{{{t}^{2}}+1}dt}\]\[=\int{2dt-\int{\frac{2dt}{{{t}^{2}}+1}}}\] \[=2t-2{{\tan }^{-1}}t+c=2\sqrt{{{e}^{x}}-1}-2{{\tan }^{-1}}\sqrt{{{e}^{x}}-1}+c\].
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