Evaluate:`int(x^2+1)/((x-1)^2(x+3))dx`

1.	Evaluate:`int(x^2+1)/((x-1)^2(x+3))dx`
Answer» `I=int(x^(2)+1)/((x-1)^(2)(x+3))dx` Let `int(x^(2)+1)/((x-1)^(2)(x+3))=(A)/(x-1)+(B)/((x-1)^(2))+(C)/(x+3) " " ` (1) or `x^(2)+1=A(x-1)(x+3)+B(x+3)+C(x-1)^(2) " " ` (2) Putting `x-1=0`, i.e., `x=1` in equation (2), we get `2=4B " or " B=(1)/(2).` Putting `x+3=0`, i.e., `x= -3` in equation (2), we get `10=16C " or " C=(5)/(8).` Equating the coefficients of `x^(2)` on both the sides of the identity of equation (2), we get `I=A+C " or " A=1-C=1-(5)/(8)=(3)/(8)` Substituting the values of A, B in equation (1), we get `(x^(2)+1)/((x-1)^(2)(x+3))=(3)/(8)(1)/(x-1)+(1)/(2)(1)/((x-1)^(2))+(5)/(8)(x+3)` or `I=(3)/(8)int(1)/(x-1)dx+(1)/(2)int(1)/((x-1)^(2))dx+(5)/(8)int(1)/(x+3)dx` `=(3)/(8)log\|x-1\|-(1)/(2(x-1))+(5)/(8)log\|x+3\|+C`

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Evaluate:`int(x^2+1)/((x-1)^2(x+3))dx`

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