Integral Calculus Question 66
Question: $ \int_{{}}^{{}}{\frac{dx}{x[{{(\log x)}^{2}}+4\log x-1]}}= $
Options:
A) $ \frac{1}{2\sqrt{5}}\log [ \frac{\log x+2-\sqrt{5}}{\log x+2+\sqrt{5}} ]+c $
B) $ \frac{1}{\sqrt{5}}\log [ \frac{\log x+2-\sqrt{5}}{\log x+2+\sqrt{5}} ]+c $
C) $ \frac{1}{2\sqrt{5}}\log [ \frac{\log x+2+\sqrt{5}}{\log x+2-\sqrt{5}} ]+c $
D) $ \frac{1}{\sqrt{5}}\log [ \frac{\log x+2+\sqrt{5}}{\log x+2-\sqrt{5}} ]+c $
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Answer:
Correct Answer: A
Solution:
Put $ \log x=t\Rightarrow \frac{1}{x},dx=dt, $ then $ \int_{{}}^{{}}{\frac{dx}{x[{{(\log x)}^{2}}+4\log x-1]}}=\int_{{}}^{{}}{\frac{dt}{t^{2}+4t-1}} $ $ =\int_{{}}^{{}}{\frac{dt}{{{(t+2)}^{2}}-{{(\sqrt{5})}^{2}}}=\frac{1}{2\sqrt{5}}\log [ \frac{t+2-\sqrt{5}}{t+2+\sqrt{5}} ]} $ $ =\frac{1}{2\sqrt{5}}\log [ \frac{\log x+2-\sqrt{5}}{\log x+2+\sqrt{5}} ]+c $ .
BETA
