A) \[2[\sqrt{{{e}^{x}}-1}-{{\tan }^{-1}}\sqrt{{{e}^{x}}-1}]+c\]
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) \[2[\sqrt{{{e}^{x}}-1}+{{\tan }^{-1}}\sqrt{{{e}^{x}}-1}]+c\]
E) \[2[\sqrt{{{e}^{x}}-1}-{{\tan }^{-1}}\sqrt{{{e}^{x}}+1}]+c\]
Correct Answer: A
Solution :
Let \[I=\int{\sqrt{{{e}^{x}}-1}}dx=\int{\frac{\sqrt{{{e}^{x}}-1}\,{{e}^{x}}}{1+{{(\sqrt{{{e}^{x}}-1})}^{2}}}}dx\] Put \[\sqrt{{{e}^{x}}-1}=t\Rightarrow {{e}^{x}}-1={{t}^{2}}\] \[\Rightarrow \] \[{{e}^{x}}dx=2t\,dt\] \[\therefore \] \[I=2\int{\frac{{{t}^{2}}dt}{1+{{t}^{2}}}}=2\int{\left( \frac{1+{{t}^{2}}}{1+{{t}^{2}}} \right)}dt\] \[-2\int{\frac{1}{1+{{t}^{2}}}}dt\] \[=2[t-{{\tan }^{-1}}t]+c\] \[=2[\sqrt{{{e}^{x}}-1}-{{\tan }^{-1}}\sqrt{{{e}^{x}}-1}]+c\]
You need to login to perform this action.
You will be redirected in
3 sec
