Integral Calculus Question 301
Question: $ \int_{{}}^{{}}{\frac{\log (x+\sqrt{1+x^{2}})}{\sqrt{1+x^{2}}}\ dx=} $
Options:
A) $ \frac{1}{2}{{[\log (x+\sqrt{1+x^{2}})]}^{2}}+c $
B) $ \log {{(x+\sqrt{1+x^{2}})}^{2}}+c $
C) $ \log (x+\sqrt{1+x^{2}})+c $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Put $ \log (x+\sqrt{1+x^{2}})=t\Rightarrow \frac{1}{\sqrt{1+x^{2}}},dx=dt, $ then $ \int_{{}}^{{}}{\frac{\log (x+\sqrt{1+x^{2}})}{\sqrt{1+x^{2}}},dx}=\int_{{}}^{{}}{t,dt} $ $ =\frac{1}{2}{{[ \log (x+\sqrt{1+x^{2}}) ]}^{2}}+c $ .
BETA
