Definite Integration Question 266
Question: The value of $ \int_1^{2}{\frac{dx}{x(1+x^{4})}} $ is
[MP PET 2004]
Options:
A) $ \frac{1}{4}\log \frac{17}{32} $
B) $ \frac{1}{4}\log \frac{17}{2} $
C) $ \log \frac{17}{2} $
D) $ \frac{1}{4}\log \frac{32}{17} $
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Answer:
Correct Answer: D
Solution:
$ \int_1^{2}{\frac{dx}{x(1+x^{4})}=\int_1^{2}{\frac{dx}{x^{5}( 1+\frac{1}{x^{4}} )}}} $
Put $ ( 1+\frac{1}{x^{4}} )=z\Rightarrow \frac{-4}{x^{5}}dx=dz $
therefore $ \frac{-1}{4}\int_2^{17/16}{\frac{dz}{z}=[ \frac{-1}{4}\log z ]_2^{17/16}} $ = $ \frac{1}{4}\log 2-\frac{1}{4}\log \frac{17}{16} $
therefore $ I=\frac{1}{4}\log ( \frac{32}{17} ) $ .
BETA
