How do you integrate #(2x)/((x-1)(x+1))# using partial fractions?

Answer 1

#ln|x+1|+ln|x-1|+C #where C is a constant

The given expression can be written as partial sum of fractions:

#(2x)/((x+1)(x-1))=1/(x+1)+1/(x-1)#

Now let's integrate :

#int(2x)/((x+1)(x-1))dx#
#int1/(x+1)+1/(x-1)dx#
#int1/(x+1)dx+int1/(x-1)dx#
#int(d(x+1))/(x+1)+int(d(x-1))/(x-1)#
#ln|x+1|+ln|x-1|+C #where C is a constant
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Answer 2

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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Answer from HIX Tutor

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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