Question
Evaluate the following integrals:$\int^\limits1_0\frac{1-\text{x}^2}{\text{x}^4+\text{x}^2+1}\text{ dx}$
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Answer
Let $\text{I}=\int^\limits1_0\frac{1-\text{x}^2}{\text{x}^4+\text{x}^2+1}\text{ dx}$$=-\int\frac{\text{x}^2-1}{\text{x}^4+\text{x}^2+1}\text{ dx}$
$=-\int\frac{1-\frac{1}{\text{x}^2}}{\text{x}^2+1+\frac{1}{\text{x}^2}}\text{ dx}$
$=-\int\frac{1-\frac{1}{\text{x}^2}}{\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-1}\text{ dx}$
Let, $\text{x}+\frac{1}{\text{x}}=\text{t}$$\Big(1-\frac{1}{\text{x}^2}\Big)\text{dx}=\text{dt}$
Then integral becames,$\text{I}=-\int\frac{1}{\text{t}^2-1}\text{dt}$
$=-\frac{1}{2}\log\Big|\frac{\text{t}-1}{\text{t}+1}\Big|$
$=\frac{1}{2}\log\Big|\frac{\text{t}+1}{\text{t}-1}\Big|$
$=\frac{1}{2}\log\Big|\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big|$
i,e., $\int\frac{1-\text{t}^2}{\text{x}^4+\text{x}^2+1}\text{ dx}=\frac{1}{2}\log\Big|\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big|$$\int^\limits1_0\frac{1-\text{x}^2}{\text{x}^4+\text{x}^2+1}\text{ dx}=\Big[\frac{1}{2}\log\Big|\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big|\Big]^1_0$
$=\frac{1}{2}\log3$
$=-\int\frac{1-\frac{1}{\text{x}^2}}{\text{x}^2+1+\frac{1}{\text{x}^2}}\text{ dx}$
$=-\int\frac{1-\frac{1}{\text{x}^2}}{\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-1}\text{ dx}$
Let, $\text{x}+\frac{1}{\text{x}}=\text{t}$$\Big(1-\frac{1}{\text{x}^2}\Big)\text{dx}=\text{dt}$
Then integral becames,$\text{I}=-\int\frac{1}{\text{t}^2-1}\text{dt}$
$=-\frac{1}{2}\log\Big|\frac{\text{t}-1}{\text{t}+1}\Big|$
$=\frac{1}{2}\log\Big|\frac{\text{t}+1}{\text{t}-1}\Big|$
$=\frac{1}{2}\log\Big|\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big|$
i,e., $\int\frac{1-\text{t}^2}{\text{x}^4+\text{x}^2+1}\text{ dx}=\frac{1}{2}\log\Big|\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big|$$\int^\limits1_0\frac{1-\text{x}^2}{\text{x}^4+\text{x}^2+1}\text{ dx}=\Big[\frac{1}{2}\log\Big|\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big|\Big]^1_0$
$=\frac{1}{2}\log3$
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