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How do you integrate #int 1/(n(n+2))# using partial fractions?

Answer 1

#int 1/(n(n+2)) dn = 1/2 ln abs(n) - 1/2 ln abs(n+2) + C#

We are searching for a partical fraction decomposition of the following form, as the denominator has already been factorized into separate linear factors:

#1/(n(n+2)) = A/n + B/(n+2)#

#=(A(n+2)+Bn)/(n(n+2))#

#=((A+B)n+2A)/(n(n+2))#

Coefficients are equated to yield:

#{(A+B=0), (2A=1) :}#

Hence:

#{(A=1/2),(B=-1/2):}#

So:

#int 1/(n(n+2)) dn = int (1/(2n) - 1/(2(n+2))) dn = 1/2 ln abs(n) - 1/2 ln abs(n+2) + C#

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Answer 2

Christopher Agresta

To integrate ( \int \frac{1}{n(n+2)} ) using partial fractions, we first express the integrand as a sum of simpler fractions.

( \frac{1}{n(n+2)} = \frac{A}{n} + \frac{B}{n+2} )

Now, we find the values of ( A ) and ( B ) by multiplying both sides by the common denominator ( n(n+2) ):

( 1 = A(n+2) + Bn )

Expanding and equating coefficients, we get two equations:

For ( n ): ( 0 = A + B ) For ( n+2 ): ( 1 = 2A )

Solving these equations yields ( A = \frac{1}{2} ) and ( B = -\frac{1}{2} ).

So, ( \frac{1}{n(n+2)} = \frac{1}{2n} - \frac{1}{2(n+2)} ).

Now, you can integrate each term separately.

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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.