Integrate the rational function in exercise: x (x^2+1)(x-1) - Vidyadip

Question

Integrate the rational function in exercise:
$\frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}$

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Answer

$\frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}=\frac{\text{Ax}+\text{B}}{\text{x}^2+1}+\frac{\text{C}}{\text{x}-1}\dots(\text{i)}$
⇒ x = (Ax + B) (x - 1) + C(x² + 1)
⇒ x = Ax² - Ax + Bx - B + Cx² + C
Comparing coefficients of x², A + C = 0 ....(ii)
Comparing coefficients of x, –A + B = 1 ....(iii)
Comparing constant terms, –B + C = 0 ….(iv)
Solving eq. (ii), (iii) and (iv), we get
$\text{A}=\frac{-1}{2},\ \text{B}=\frac{1}{2} \ \text{and} \ \text{C}=\frac{1}{2}$
Putting the values of A, B and C in eq. (i),
$\frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}=\frac{\frac{-1}{2}\text{x}+\frac{1}{2}}{\text{x}^2+1}+\frac{\frac{1}{2}}{\text{x}-1}$
$\Rightarrow \ \frac{\text{x}}{(\text{x}^2+1)}=\frac{-1}{2}.\frac{\text{x}}{\text{x}^2+1}+\frac{1}{2}.\frac{1}{\text{x}^2+1}+\frac{1}{2}.\frac{1}{\text{x}-1}=\frac{-1}{4}.\frac{2\text{x}}{\text{x}^2+1}+\frac{1}{2}.\frac{1}{\text{x}^2+1}+\frac{1}{2}.\frac{1}{\text{x}-1}$
$\Rightarrow \ \frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}\text{dx}=\frac{-1}{4}\int\frac{2\text{x}}{\text{x}^2+1}\text{dx}+\frac{1}{2}\int\frac{1}{\text{x}^2+1}\text{dx}+\frac{1}{2}\int\frac{1}{\text{x}-1}\text{dx}$
$\Rightarrow \ \frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}\text{dx}=\frac{-1}{4}\text{log}|\text{x}^2+1|+\frac{1}{2}\tan^{-1}\text{x}+\frac{1}{2}\text{log}|\text{x}-1|+\text{c}$
$\Rightarrow \ \frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}=\frac{-1}{4}\text{log}(\text{x}^2+1)+\frac{1}{2}\tan^{-1}\text{x}+\frac{1}{2}\text{log}|\text{x}-1|+\text{c}$

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