Integral Calculus Question 149
Question: If $ \int{{\log_{e}}( \sqrt{1-x}+\sqrt{1+x} )dx} $ $ =x{\log_{e}}( \sqrt{1-x}+\sqrt{1+x} )+g(x)+C $ . Then $ g(x)= $
Options:
A) $ x-{{\sin }^{-1}}x $
B) $ {{\sin }^{-1}}x-x $
C) $ x+{{\sin }^{-1}}x $
D) $ {{\sin }^{-1}}x-x^{2} $
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Answer:
Correct Answer: B
Solution:
[b] $ I=\int{{\log_{e}}( \sqrt{1-x}+\sqrt{1+x} ).1dx} $ Integrating by parts taking 1 as the second function. $ I=\log ( \sqrt{1-x}+\sqrt{1+x} )x-\int{\frac{1}{\sqrt{1-x}+\sqrt{1+x}}} $ $ -\frac{1}{2\sqrt{1-x}}+\frac{1}{2\sqrt{1+x}} dx $ $ =x\log ( \sqrt{1-x}+\sqrt{1+x} ) $ $ =\frac{1}{2}\int{\frac{\sqrt{1-x}-\sqrt{1+x}}{\sqrt{1-x}+\sqrt{1+x}}.\frac{1}{\sqrt{1+x^{2}}}.xdx} $ $ =x\log (\sqrt{1-x}+\sqrt{1+x}) $ $ -\frac{1}{2}\int{\frac{(1-x)+(1+x)-2\sqrt{1-x^{2}}}{(1-x)-(1+x)}.\frac{1}{\sqrt{1-x^{2}}}.xdx} $ $ =x\log ( \sqrt{1-x}+\sqrt{1+x} )-\frac{1}{2}\int{\frac{\sqrt{1-x^{2}}-1}{\sqrt{1-x^{2}}}dx} $ $ =x\log ( \sqrt{1-x}+\sqrt{1+x} )-\frac{1}{2}[ \int{1dx-}\int{\frac{1}{\sqrt{1-x^{2}}}dx} ] $ $ =x\log ( \sqrt{1-x}+\sqrt{1+x} )+\frac{1}{2}[ {{\sin }^{-1}}x-x ]+C $
$ \therefore f(x)=x,g(x)={{\sin }^{-1}}x-x $
BETA
