Integral Calculus Question 20
Question: $ \int_{{}}^{{}}{\frac{dx}{1+x+x^{2}+x^{3}}=} $
[MP PET 1991]
Options:
A) $ \log | \sqrt{1+x}|-\frac{1}{2} \log |\sqrt{1+x^{2}}|+\frac{1}{2}{{\tan }^{-1}}x+c $
B) $ \log |\sqrt{1+x}|-\log |\sqrt{1+x^{2}}|+{{\tan }^{-1}}x+c $
C) $ \log |\sqrt{1+x^{2}}|- \log |\sqrt{1+x}|+\frac{1}{2}{{\tan }^{-1}}x+c $
D) $ \log |\sqrt{1+x}|+{{\tan }^{-1}}x+\log |\sqrt{1+x^{2}}|+c $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\frac{dx}{1+x+x^{2}+x^{3}}=\int_{{}}^{{}}{\frac{dx}{(1+x)(1+x^{2})}}} $ $ =\frac{1}{2}\int_{{}}^{{}}{\frac{1}{1+x^{2}},dx}+\frac{1}{2}\int_{{}}^{{}}{\frac{1}{1+x},dx}-\frac{1}{2}\int_{{}}^{{}}{\frac{x}{1+x^{2}},dx} $ $ =\frac{1}{2}{{\tan }^{-1}}x+\log |\sqrt{1+x}-\frac{1}{2}| \log| \sqrt{1+x^{2}}|+c $ .
BETA
