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April 24, 2014

April 24, 2014

Posted by **Gabrielle** on Tuesday, September 20, 2011 at 8:31pm.

InT(-x^3+9x^2-3x+2)/(x^4-2x^3)

I got ((1-2x)/(2x^2))+3*ln(2-x)-4*ln(x)

but apparently that's not the answer...

- Calculus -
**Count Iblis**, Tuesday, September 20, 2011 at 9:53pmThe answer is correct, except that you need to take te absolute value of all the arguments of the logarithms, so

ln(x) should be ln|x|

ln(2-x) should be ln|2-x| = ln|x-2|

And there is, of course, an integration constant.

The following method is easier, doesn't involv solving equations, and it is more error proof than the traditional method:

The integrand is:

f(x) = (-x^3+9x^2-3x+2)/(x^4-2x^3)

The numerator can be factored as:

x^4-2x^3 = x^3 (x-2)

f(x) thus has singularities at x = 2 and x = 0. The series expansion of f(x) around these points starts with singular terms. Of the expansion around x = 2, only the first term is singular:

f(x) = (-x^3+9x^2-3x+2)/[x^3 (x-2)] =

1/(x-2) (-x^3+9x^2-3x+2)/x^3

Then we need to expand

(-x^3+9x^2-3x+2)/x^3

in powers of (x - 2 ), but we only need the constant term as higher powers pultiplied by 1/(x-2) are non-singular. But this is simply the value at x = 2, which is 3.

So, the singular part of f(x) near x = 2 is

s(x) = 3/(x-2)

Similarly, f(x) has a singularity at x = 0, and we can compute the singular part of the expansin there, which will be some function r(x). Let's skip this part and see how that would solve the problem. Consider the function

g(x)= f(x) - s(x) - r(x)

Clearly g(x) is a rational function, but it doesn't have any singularities, so it must be a polynomial. Now the degree of the numerator of f(x) is less than the degree of the denominator so for x to infinity, f(x) tends to zero.

s(x) and r(x) also tend to zero for x to infinity.

This means that the limit if x to infinity of g(x) is zero. But the x to infinity limit of a polynomial in x can only be zero if the polynomial is identically zero. So, g(x) = 0, and we have:

f(x) - s(x) - r(x) = 0 ---->

f(x) = s(x) + r(x)

This is then the desired partial fraction expansion, except that we haven't calculated r(x) yet.

We can avoid computing r(x) via the expansion around x = 0 (which is more work than computing s(x), because the expansion starts with 1/x^3, so we need to expand the factor multiplying /x^3 to second order in x to obtain all singular terms) as follows. We can write:

r(x) = f(x) - s(x) =

(-x^3+9x^2-3x+2)/[x^3 (x-2)] - 3/(x-2) =

(-4x^3+9x^2-3x+2)/[x^3 (x-2)]

Te numerator must be divisible by (x-2), because r(x) is not singular there:

-4x^3+9x^2-3x+2 =

-4 x^3 + 8 x^2 - 8 x^2 + 9x^2-3x+2 =

-4 x^3 + 8 x^2 + x^2 - 3 x + 2 =

-4 x^2 (x-2) + (x-1)(x-2) =

(x-2) (-4 x^2 + x-1)

So, we have:

r(x) = (-4 x^2 + x-1)/x^3 =

-1/x^3 + 1/x^2 -4/x

Therefore:

f(x) = s(x) + r(x) =

3/(x-2) -1/x^3 + 1/x^2 -4/x

And integrating yields the same answer.

So, we didn't need to solve any equations, and mistakes in computations don't propagate. E.g. we don't need to check whether s(x) has been correctly computed, because using the s(x) we obtained, we computed

r(x) = f(x) - s(x) and this was of the correct form, and f(x) is then

r(x) + s(x) which we can integrate. Posible mistakes made when computing

s(x) are thus completely irrelevant.

- Calculus -
**Reiny**, Tuesday, September 20, 2011 at 10:15pmA difficult split up into partial fractions

let (-x^3+9x^2-3x+2)/(x^4-2x^3)

= A/x + B/x^2 + C/x^3 + D/(x-2)

then

Ax^2(x-2) + Bx(x-2) + C(x-2) + Dx^3 = -x^3 + 9x^2 - 3x + 2

This is an identity, so it must be true for any x.

Pick suitable values of x

let x=0 --> -2C=2

C=-1

let x = 2 --> 8D = -8+36-6+2

D = 3

so Ax^2(x-2) + Bx(x-2) -x+2 + 3x^3 = -x^3 + 9x^2 - 3x + 2

let x=1 --> -A - B + 1 + 3 = -1 +9-3+2

A+B = -3

let x = 3 --> 9A + 3B - 3+2+81 = -27+81-9+2

9A + 3B = -33

Solving these last two equations in A and B gives us

A = -4

B = 1

so (-x^3+9x^2-3x+2)/(x^4-2x^3)

= -4/x + 1/x^2 - 1/x^3 + 3/(x-2)

and the integral of that is

-4lnx - 1/x + 1/(2x^2) + 3ln(x-2) + c

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