Evaluate the following intregals: x (x^2+1)(x-1) dx - Vidyadip

Question

Evaluate the following intregals:
$\int\frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}\ \text{dx}$

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Answer

Consider the integrals
$\text{I}=\int\frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}\ \text{dx}$
Now let us separate the fraction $\frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}$
Through particle fractions.
$\frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}=\frac{\text{A}}{\text{x}-1}+\frac{\text{Bx}+\text{C}}{\text{x}^2+1}$
$\Rightarrow\frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}=\frac{\text{A}(\text{x}^2+1)+(\text{Bx}+\text{C})(\text {x}-1)}{(\text{x}^2+1)(\text{x}-1)}$
$\Rightarrow\text{x}=\text{A}(\text{x}^2+1)+(\text{Bx}+\text{C})(\text{x}-1)$
Compairing the coefficient, we have,
A + B = 0, -B + C = 1 and A - C = 0
Solving the equations we, get,
$\Rightarrow\text{A}=\frac{1}{2},\text{B}=-\frac{1}{2}\text{ and }\text{C}=\frac{1}{2}$
$\Rightarrow\frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}=\frac{\text{A}}{\text{x}-1}+\frac{\text{Bx}+\text{C}}{\text{x}^2+1}$
$\Rightarrow\frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}=\frac{1}{2}\times\frac{1}{\text{x}-1}-\frac{1}{2}\times\frac{\text{x}-1}{\text{X}^2+1}$
$\Rightarrow\frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}=\frac{1}{2(\text{x}-1)}-\frac{\text{x}}{2(\text{x}^2+1)}+\frac{1}{2(\text{x}^2+1)}$
Thus, we have,
$=\int\frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}\ \text{dx}$
$=\int\Big[\frac{1}{2(\text{x}-1)}-\frac{\text{x}}{2(\text{x}^2+1)}+\frac{1}{2(\text{x}^2+1)}\Big]\text{dx}$
$=\int\frac{\text{dx}}{2(\text{x}-1)}-\int\frac{\text{xdx}}{2(\text{x}^2+1)}+\int\frac{\text{dx}}{2(\text{x}^2+1)}$
$=\int\frac{1}{2}\frac{\text{dx}}{(\text{x}-1)}-\int\frac{1}{2}\frac{\text{xdx}}{(\text{x}^2+1)}+\int\frac{1}{2}\frac{\text{dx}}{(\text{x}^2+1)}$
$=\frac{1}{2}\int\frac{\text{dx}}{(\text{x}-1)}-\frac{1}{2}\times\frac{1}{2}\int\frac{2\text{xdx}}{(\text{x}^2+1)}+\frac{1}{2}\int\frac{\text{dx}}{(\text{x}^2+1)}$
$=\frac{1}{2}\log|\text{x}-1|-\frac{1}{4}\log(\text{x}^2+1)+\frac{1}{2}\tan^{-1}\text{x}+\text{C}$

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