How do you integrate #(2x)/((x-1)(x+1))# using partial fractions?
The given expression can be written as partial sum of fractions:
Now let's integrate :
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To integrate ( \frac{2x}{(x-1)(x+1)} ) using partial fractions, first, we express the fraction as a sum of simpler fractions:
[ \frac{2x}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1} ]
We then solve for ( A ) and ( B ) by equating numerators:
[ 2x = A(x+1) + B(x-1) ]
Now, we can solve for ( A ) and ( B ) by substituting suitable values for ( x ). For instance, setting ( x = 1 ) gives:
[ 2 = 2A ]
And setting ( x = -1 ) gives:
[ -2 = -2B ]
Thus, ( A = 1 ) and ( B = -1 ).
Now, we can rewrite the original fraction as:
[ \frac{1}{x-1} - \frac{1}{x+1} ]
Integrating each term separately gives:
[ \int \frac{1}{x-1} , dx - \int \frac{1}{x+1} , dx ]
[ = \ln|x-1| - \ln|x+1| + C ]
Therefore, the integral of ( \frac{2x}{(x-1)(x+1)} ) using partial fractions is ( \ln|x-1| - \ln|x+1| + C ), where ( C ) is the constant of integration.
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When evaluating a one-sided 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 one-sided 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 one-sided 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 one-sided 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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