How do I find the integral #int1/((w-4)(w+1))dw# ?

Answer 1
#=1/5ln((w-4)/(w+1))+c#, where #c# is a constant

Explanation :

This type of question usually solve by using Partial Fractions,

#1/((w-4)(w+1))#, it can be written as
#1/((w-4)(w+1))=A/(w-4)+B/(w+1)#
multiplying by #(w-4)(w+1)# on both sides, we get
#1=A(w+1)+B(w-4)#
#1=(A+B)w+(A-4B)#
Now comparing coefficient of #w# and constants both sides, we get
#A+B=0# #=># #A=-B# ...........#(i)# #A-4B=1# ..............#(ii)#
Substituting value of #A# from #(i)# to #(ii)#, we get
#-5B=1# #=># #B=-1/5#
from #B#, we can easily calculate #A#, which will be #1/5#

Now,

#1/((w-4)(w+1))=1/(5(w-4))-1/(5(w+1))#
Integrating both side with respect to #w#,
#int1/((w-4)(w+1))dw=int1/(5(w-4))dw-int1/(5(w+1))dw#
#=1/5(ln(w-4)-ln(w+1))+c#, where #c# is a constant
#=1/5ln((w-4)/(w+1))+c#, where #c# is a constant
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Answer 2

To find the integral of 1/((w-4)(w+1))dw, you can use partial fraction decomposition. After decomposing the expression into partial fractions, you integrate each term separately. The integral will involve natural logarithms and constant terms.

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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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