How do you integrate #int 1/ [(x1) ( x+ 1) ^2] # using partial fractions?
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To integrate ( \frac{1}{(x1)(x+1)^2} ) using partial fractions:

Write ( \frac{1}{(x1)(x+1)^2} ) as a sum of partial fractions: ( \frac{1}{(x1)(x+1)^2} = \frac{A}{x1} + \frac{B}{x+1} + \frac{C}{(x+1)^2} )

Clear the denominators by multiplying both sides by ( (x1)(x+1)^2 ): ( 1 = A(x+1)^2 + B(x1)(x+1) + C(x1) )

Expand and simplify the equation: ( 1 = A(x^2 + 2x + 1) + B(x^2  x + x  1) + C(x1) ) ( 1 = Ax^2 + 2Ax + A + Bx^2  B + Cx  C + Cx  C ) ( 1 = (A + B)x^2 + (2A + 2B + C)x + (A  B  C) )

Equate coefficients: ( A + B = 0 ) (for ( x^2 ) terms) ( 2A + 2B + C = 0 ) (for ( x ) terms) ( A  B  C = 1 ) (for constant terms)

Solve the system of equations for ( A ), ( B ), and ( C ):
From the first equation, ( B = A ).
Substitute ( B = A ) into the second equation: ( 2A  2A + C = 0 ) gives ( C = 0 ).
Substitute ( B = A ) and ( C = 0 ) into the third equation: ( A + A  0 = 1 ) gives ( A = \frac{1}{2} ).
Since ( B = A ), then ( B = \frac{1}{2} ).

Rewrite the original integral with the partial fraction decomposition: ( \int \frac{1}{(x1)(x+1)^2} , dx = \int \left(\frac{1/2}{x1}  \frac{1/2}{x+1} + \frac{0}{(x+1)^2}\right) , dx )

Integrate each term: ( \frac{1}{2} \lnx1  \frac{1}{2} \lnx+1 + C )

Combine the terms and add the constant of integration ( C ): ( \frac{1}{2} \lnx1  \frac{1}{2} \lnx+1 + C )
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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