Integral Calculus Question 343
Question: $ \int_{{}}^{{}}{\frac{x+1}{\sqrt{1+x^{2}}}dx}= $
[MP PET 1991]
Options:
A) $ \sqrt{1+x^{2}}+{{\tan }^{-1}}x+c $
B) $ \sqrt{1+x^{2}}-\log {x+\sqrt{1+x^{2}}}+c $
C) $ \sqrt{1+x^{2}}+\log {x+\sqrt{1+x^{2}}}+c $
D) $ \sqrt{1+x^{2}}+\log (\sec x+\tan x)+c $
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Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{\frac{x+1}{\sqrt{x^{2}+1}},dx=\int_{{}}^{{}}{\frac{x}{\sqrt{x^{2}+1}},dx+\int_{{}}^{{}}{\frac{1}{\sqrt{x^{2}+1}},dx}}} $ Put $ x^{2}+1=t\Rightarrow 2x,dx=dt $ , then it reduce to $ \frac{1}{2}\int_{{}}^{{}}{\frac{dt}{{t^{1/2}}}+\int_{{}}^{{}}{\frac{1}{\sqrt{x^{2}+1}}}},dx=\frac{1}{2}.2.{t^{1/2}}+\log (x+\sqrt{x^{2}+1})+c $ $ ={{(x^{2}+1)}^{1/2}}+\log (x+\sqrt{x^{2}+1})+c. $
BETA
