Evaluate the following intregals: x^3 (x-1)(x-2)(x-3) dx - Vidyadip

Question

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

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Answer

Let $\text{I}=\int\frac{\text{x}^3}{(\text{x}-1)(\text{x}-2)(\text{x}-3)}\ \text{dx}$
$=\int1+\frac{6\text{x}^2-9\text{x}+6}{(\text{x}-1)(\text{x}-2)(\text{x}-3)}\ \text{dx}$
Let $\frac{6\text{x}^2-9\text{x}+6}{(\text{x}-1)(\text{x}-2)(\text{x}-3)}=\frac{\text{A}}{\text{x}-1}+\frac{\text{B}}{\text{x}-2}+\frac{\text{C}}{\text{x}-3}$
$\text{x}\Rightarrow6\text{x}^2-11+6=\text{A}(\text{x}-2)(\text{x}-3)\\+\text{B}(\text{x}-1)(\text{x}-3)+\text{C}(\text{x}-1)(\text{x}-2)$
put x = 1
$\Rightarrow1=2\text{A}\Rightarrow\text{A}=\frac{1}{2}$
put x = 2
$\Rightarrow8=-\text{B}\Rightarrow\text{B}=-8$
put x = 3
$\Rightarrow27=2\text{C}\Rightarrow\text{C}=\frac{27}{2}$
Thus,
$\text{I}=\int\text{dx}+\frac{1}{2}\int\frac{\text{dx}}{\text{x}-1}-8\int\frac{\text{dx}}{\text{x}-2}+\frac{27}{2}\int\frac{\text{dx}}{\text{x}-3}$
$=\text{x}+\frac{1}{2}\log|\text{x}-1|-8\log|\text{x}-2|+\frac{27}{2}\log|\text{x}-3|+\text{C}$
Hence,
$\text{I}=\text{x}+\frac{1}{2}\log|\text{x}-1|-8\log|\text{x}-2|+\frac{27}{2}\log|\text{x}-3|+\text{C}$

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