How do you integrate #(x^3-4x-10)/(x^2-x-6)# using partial fractions?
1 Answer
Explanation:
The first step is to factor the denominator of the function
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since the factors are linear then the coefficients of the partial fractions will be constants , say A and B. Writing the function in terms of it's partial fractions.
multiplying through by (x-3)(x+2)
We now have to find the values of A and B .Note that if x = -2 the term with A will be zero and if x = 3 the term with B will be zero. We can make use of this fact in finding A and B.
let x = -2 in (1) :- 10 = -5B #rArrcolor(blue)' B = 2'#
let x = 3 in (1) : 5 = 5A #rArr color(blue)' A = 1 '#
Integral becomes : #int(dx)/(x-3) + int(2dx)/(x+2) #
Related topic
Related questions
Doo 2 3 3 X 2
How do you factor # x^4 + x^3 + x^2 + x + 1#?
1 Answer
Explanation:
This quartic has four zeros, which are the non-Real Complex #5# th roots of #1# , as we can see from:
So if we wanted to factor this polynomial as a product of linear factors with Complex coefficients then we could write:
A cleaner algebraic approach is to notice that due to the symmetry of the coefficients, if #x=r# is a zero of #x^4+x^3+x^2+x+1# , then #x=1/r# is also a zero.
Hence there is a factorisation in the form:
So let's look for a factorisation:
3 X 2 2x X 3
![Doo 2 3 3 X 2 Doo 2 3 3 X 2](https://hi-static.z-dn.net/files/ded/8aee0663a479dda7b053d15eb69ac98c.jpg)
Equating coefficients we find:
3 5 X 2 3 Fraction
Substituting #b=-1/a# in #a+b=1# we get:
Hence:
Using the quadratic formula, we can deduce:
Since our derivation was symmetric in #a# and #b# , one of these roots can be used for #a# and the other for #b# , to find:
If we want to factor further, use the quadratic formula on each of these quadratic factors to find the linear factors with Complex coefficients.