Integral Calculus Question 56
Question: $ \int_{{}}^{{}}{\frac{dx}{2x^{2}+x+1}}\ $ equals
[RPET 1997]
Options:
A) $ \frac{1}{\sqrt{7}}{{\tan }^{-1}}( \frac{4x+1}{\sqrt{7}} )+c $
B) $ \frac{1}{2\sqrt{7}}{{\tan }^{-1}}( \frac{4x+1}{\sqrt{7}} )+c $
C) $ \frac{1}{2}{{\tan }^{-1}}( \frac{4x+1}{\sqrt{7}} )+c $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
$ I=\int_{{}}^{{}}{\frac{dx}{2x^{2}+x+1}}=\int_{{}}^{{}}{\frac{dx}{2( x^{2}+\frac{x}{2}+\frac{1}{2} )}} $ $ =\frac{1}{2}\int_{{}}^{{}}{\frac{dx}{x^{2}+\frac{x}{2}+\frac{1}{16}-\frac{1}{16}+\frac{1}{2}}}=\frac{1}{2}\int_{{}}^{{}}{\frac{dx}{{{( x+\frac{1}{4} )}^{2}}+{{( \frac{\sqrt{7}}{4} )}^{2}}}} $ $ =\frac{1}{2}\frac{1}{\frac{\sqrt{7}}{4}}{{\tan }^{-1}}\frac{[x+(1/4)]}{\sqrt{7}/4} $ $ =\frac{2}{\sqrt{7}}{{\tan }^{-1}}\frac{(4x+1)}{\sqrt{7}}+C $ .
BETA
