Question
Evaluate the following integrals:
$\int\frac{\text{x}}{\text{x}^4-\text{x}^2+1}\text{dx}$
✓
Answer
$\int\frac{\text{x dx}}{\text{x}^4-\text{x}^2+1}$
Let $\text{x}^2=\text{t}$
$\Rightarrow2\text{x dx = dt}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$
Now, $\int\frac{\text{x dx}}{\text{x}^4-\text{x}^2+1}$
$ =\frac{1}{2}\int\frac{\text{dt}}{\text{t}^2-\text{t}+1}$
$=\frac{1}{2}\int\frac{\text{dt}}{\text{t}^2-\text{t}+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2+1}$
$=\frac{1}{2}\int\frac{\text{dt}}{\big(\text{t}-\frac{1}{2}\big)^2+\frac{3}{4}}$
$=\frac{1}2{}\int\frac{\text{dt}}{\Big(\text{t}-\frac{1}{2}\Big)^2+\Big(\frac{\sqrt{3}}{2}\Big)^2}$
$=\frac{1}{2}\times\frac{2}{\sqrt{3}}\tan^{-1}\Bigg(\frac{\text{t}-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\Bigg)+\text{C}$
$=\frac{1}{\sqrt{3}}\tan^{-1}\Big(\frac{2\text{t}-1}{\sqrt{3}}\Big)+\text{C}$
$=\frac{1}{\sqrt{3}}\tan^{-1}\Big(\frac{2\text{x}^2-1}{\sqrt{3}}\Big)+\text{C}$
Let $\text{x}^2=\text{t}$
$\Rightarrow2\text{x dx = dt}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$
Now, $\int\frac{\text{x dx}}{\text{x}^4-\text{x}^2+1}$
$ =\frac{1}{2}\int\frac{\text{dt}}{\text{t}^2-\text{t}+1}$
$=\frac{1}{2}\int\frac{\text{dt}}{\text{t}^2-\text{t}+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2+1}$
$=\frac{1}{2}\int\frac{\text{dt}}{\big(\text{t}-\frac{1}{2}\big)^2+\frac{3}{4}}$
$=\frac{1}2{}\int\frac{\text{dt}}{\Big(\text{t}-\frac{1}{2}\Big)^2+\Big(\frac{\sqrt{3}}{2}\Big)^2}$
$=\frac{1}{2}\times\frac{2}{\sqrt{3}}\tan^{-1}\Bigg(\frac{\text{t}-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\Bigg)+\text{C}$
$=\frac{1}{\sqrt{3}}\tan^{-1}\Big(\frac{2\text{t}-1}{\sqrt{3}}\Big)+\text{C}$
$=\frac{1}{\sqrt{3}}\tan^{-1}\Big(\frac{2\text{x}^2-1}{\sqrt{3}}\Big)+\text{C}$
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