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Mathematics
∫( log (x+√1+x2)/√1+x2)dx is equal to
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Q. $ \int{\frac{\log (x+\sqrt{1+{{x}^{2}}})}{\sqrt{1+{{x}^{2}}}}}dx $ is equal to
KEAM
KEAM 2010
Integrals
A
$ {{[\log (x+\sqrt{1+{{x}^{2}}})]}^{2}}+c $
B
$ x\log (x+\sqrt{1+{{x}^{2}}})+c $
C
$ \frac{1}{2}\log (x+\sqrt{1+{{x}^{2}}})+c $
D
$ \frac{1}{2}{{[\log (x+\sqrt{1+{{x}^{2}}})]}^{2}}+c $
E
$ \frac{x}{2}\log (x+\sqrt{1+{{x}^{2}}})+c $
Solution:
Let $ I=\int{\frac{\log (x+\sqrt{1+{{x}^{2}}})}{\sqrt{1+{{x}^{2}}}}}dx $
Put $ t=\log (x+\sqrt{1+{{x}^{2}}}) $ $ \frac{dt}{dx}=\frac{1}{x+\sqrt{1+{{x}^{2}}}}\left\{ 1+\frac{x}{\sqrt{1+{{x}^{2}}}} \right\} $
$=\frac{x+\sqrt{1+{{x}^{2}}}}{(x+\sqrt{1+{{x}^{2}}}).\sqrt{1+{{x}^{2}}}} $
$=\frac{1}{\sqrt{1+{{x}^{2}}}} $
$ \therefore $ $ \frac{dx}{\sqrt{1+{{x}^{2}}}}=dt $
$ \therefore $ $ I=\int{t}\,dt=\frac{{{t}^{2}}}{2}+c $
$=\frac{1}{2}{{[\log (x+\sqrt{1+{{x}^{2}}})]}^{2}}+c $