Integral Calculus Question 361
Question: $ \int_{{}}^{{}}{\frac{x^{2}+1}{x^{4}-x^{2}+1}\ dx=} $
[MP PET 1991]
Options:
A) $ {{\tan }^{-1}}( \frac{1+x^{2}}{x} )+c $
B) $ {{\cot }^{-1}}( \frac{1+x^{2}}{x} )+c $
C) $ {{\tan }^{-1}}( \frac{x^{2}-1}{x} )+c $
D) $ {{\cot }^{-1}}( \frac{x^{2}-1}{x} )+c $
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Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{\frac{x^{2}+1}{x^{4}-x^{2}+1},dx=\int_{{}}^{{}}{\frac{( 1+\frac{1}{x^{2}} )}{x^{2}+\frac{1}{x^{2}}-1}}} $
$ =\int_{{}}^{{}}{\frac{1+\frac{1}{x^{2}}}{{{( x-\frac{1}{x} )}^{2}}+1},dx=\int_{{}}^{{}}{\frac{dt}{t^{2}+1}}={{\tan }^{-1}}t+c} $ $ { Putting,x-\frac{1}{x}=t\Rightarrow ( 1+\frac{1}{x^{2}} ),dx=dt } $
$ ={{\tan }^{-1}}( x-\frac{1}{x} )+c={{\tan }^{-1}}( \frac{x^{2}-1}{x} )+c. $
BETA
