Evaluate the following integrals: 2x x^3-1 dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{2\text{x}}{\text{x}^3-1}\ \text{dx}$

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Answer

Let $\text{I}=\int\frac{2\text{x}}{\text{x}^3-1}\text{ dx}=\int\frac{2\text{x}}{(\text{x}-1)(\text{x}^2+\text{x}+1)}\ \text{dx}$
Now,
Let $\frac{2\text{x}}{(\text{x}-1)(\text{x}^2+\text{x}+1)}=\frac{\text{A}}{(\text{x}-1)^2}+\frac{\text{Bx}+\text{C}}{\text{x}^2+\text{x}+1}$
$\Rightarrow2\text{x}=\text{A}(\text{x}^2+\text{x}+1)+(\text{Bx}+\text{C})(\text{x}-1)$
$=(\text{A}+\text{B})\text{x}^2+(\text{A}-\text{B}+\text{C})\text{x}+(\text{A}-\text{C})$
Equating similar terms,
A + B = 0, A - B + C = 2, A - C = 0,
Solving, we get, $\text{A}=\frac{2}{3},\text{B}=-\frac{2}{3},\text{C}=\frac{2}{3}$
thus,
$\text{I}=\frac{2}{3}\int\frac{\text{dx}}{\text{x}-1}-\frac{2}{3}\int\frac{(\text{x}-1)\text{dx}}{\text{x}^2+\text{x}+1}$
$=\frac{2}{3}\int\frac{\text{dx}}{\text{x}-1}-\frac{2}{3}\int\frac{(\text{2x}-2)\text{dx}}{\text{x}^2+\text{x}+1}$
$\Rightarrow\text{I}=\frac{2}{3}\int\frac{\text{dx}}{\text{x}-1}-\frac{1}{3}\int\frac{2\text{x}+1}{\text{x}^2+\text{x}+1}\ \text{dx}+\int\frac{\text{dx}}{\text{x}^2+\text{x}+1}$
$=\frac{2}{3}\int\frac{\text{dx}}{\text{x}-1}-\frac{1}{3}\int\frac{2\text{x}+1}{\text{x}^2+\text{x}+1}\ \text{dx}+\int\frac{\text{dx}}{\Big(\text{x}+\frac{1}{2}\Big)^2+\Big(\frac{\sqrt{3}}{2}\Big)^2}$
$=\frac{2}{3}\log|\text{x}-1|-\frac{1}{3}\log|\text{x}^2+\text{x}+1|\\+\frac{2}{\sqrt{3}}\tan^{-1}\Bigg(\frac{\text{x}+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\Bigg)+\text{C}$
Hence,
$\text{I}=\frac{2}{3}\log|\text{x}-1|-\frac{1}{3}\log|\text{x}^2+\text{x}+1|\\+\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{2\text{x}+1}{\sqrt{3}}\Big)+\text{C}$

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