Integral Calculus Question 120
Question: $ \int{\frac{(x^{2}-1)}{x\sqrt{x^{4}+3x^{2}+1}}dx} $ is equal to
Options:
A) $ \log | x+\frac{1}{x}+\sqrt{x^{2}+\frac{1}{x^{2}}+3} |+C $
B) $ \log | x-\frac{1}{x}+\sqrt{x^{2}+\frac{1}{x^{2}}-3} |+C $
C) $ \log | x+\sqrt{x^{2}+3} |+C $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ \int{\frac{(x^{2}-1)}{x\sqrt{x^{4}+3x^{2}+1}}dx=\int{\frac{(x^{2}-1)}{x^{2}\sqrt{x^{2}+3+\frac{1}{x^{2}}}}dx}} $ $ =\int{\frac{( 1-\frac{1}{x^{2}} )}{\sqrt{{{( x+\frac{1}{x} )}^{2}}+1}}dx=\int{\frac{dz}{\sqrt{z^{2}+1}}}} $ $ [ Puttingx+\frac{1}{x}=z\Rightarrow ( 1-\frac{1}{x^{2}} )dx=dz ] $ $ =\log | z+\sqrt{z^{2}+1} |+C $ $ =\log | x+\frac{1}{x}+\sqrt{x^{2}+\frac{1}{x^{2}}+3} |+C $
BETA
