Integral Calculus Question 49
Question: $ \int_{{}}^{{}}{\frac{x}{x^{4}+x^{2}+1}dx} $ equal to
[MP PET 2004]
Options:
A) $ \frac{1}{3}{{\tan }^{-1}}( \frac{2x^{2}+1}{3} + c) $
B) $ \frac{1}{\sqrt{3}}{{\tan }^{-1}}( \frac{2x^{2}+1}{\sqrt{3}} ) + c$
C) $ \frac{1}{\sqrt{3}}{{\tan }^{-1}}(2x^{2}+1) + c$
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ I=\int_{{}}^{{}}{\frac{x}{x^{4}+x^{2}+1},dx=\int_{{}}^{{}}{\frac{xdx}{(x^{2}+x+1),(x^{2}-x+1)}}} $
$ I=\frac{1}{2}\int_{{}}^{{}}{\frac{dx}{x^{2}-x+1}-\frac{1}{2}\int_{{}}^{{}}{\frac{dx}{x^{2}+x+1}}} $
$ I=\frac{1}{2}\int_{{}}^{{}}{\frac{dx}{{{( x-\frac{1}{2} )}^{2}}+{{( \frac{\sqrt{3}}{2} )}^{2}}}-\frac{1}{2}\int_{{}}^{{}}{\frac{dx}{{{( x+\frac{1}{2} )}^{2}}+{{( \frac{\sqrt{3}}{2} )}^{2}}}}} $
$ I=\frac{1}{\sqrt{3}}{{\tan }^{-1}}( \frac{x-\frac{1}{2}}{\frac{\sqrt{3}}{2}} )-\frac{1}{\sqrt{3}}{{\tan }^{-1}}( \frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}} ) $
$ I=\frac{1}{\sqrt{3}}{{\tan }^{-1}}( \frac{2x^{2}+1}{\sqrt{3}} ) + c$ .
BETA
