Integral Calculus Question 10
Question: $ \int{\sqrt{e^{x}-1}}dx= $
[Kerala (Engg.) 2005]
Options:
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 $
Show Answer
Answer:
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{2t^{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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