Integration by long division problem

[the7joker7]

Homework Statement

Consider the indefinite integral of (4 x^3+4 x^2-96 x -100)/(x^2-25)dx

Then the integrand decomposes into the form

ax + b + c/(x - 5) + d/(x + 5)

Find a, b, c, and d.

Then find the integral of the function.

The Attempt at a Solution

Using long division, I got this far...

\frac{4x^3 + 4x^2 - 96x - 100}{x^2 - 25}

=

4x + 4 + c/(x - 5) + d/(x + 5)

It'd be pretty hard to show how I got a and b, but I'm pretty positive that's correct. I just can't find C or D.

So then, I thought I was supposed to put my remainder after dividing over (x - 5)(x + 5). My remainder is 4x, so I went.

4x/((x - 5)(x + 5))

Which produces

c + d = 4.

I need one more equation to solve for it though. Help?

 

[tiny-tim]

… just multiply …

4x/((x - 5)(x + 5))

Which produces

c + d = 4.

I need one more equation to solve for it though.

Hi the7joker7!

c/(x - 5) + d/(x + 5) = 4x/(x - 5)(x + 5);

just multiply both sides by (x - 5)(x + 5).

 

[the7joker7]

So c + d = 4 is correct?

So that gets me to...

C(x + 5) + D(x - 5) = 4x

Cx + 5c + dx + 5d = 4x

4x = (c + d)x + (5c + 5d)

Again, produces c + d = 4

So that doesn't really help. =/

 

[cristo]

Basically, you want to solve this equation:
\frac{4x^3+4x^2-96x-100}{x^2-25}=ax+b+\frac{c}{x-5}+\frac{d}{x+5}

Now, the way I would do it would be to multiply up the right hand side, and put it all over the common denominator (x-5)(x+5). Then, you can compare the numerator of the LHS to the new numerator on the RHS. Compare coefficients: you correctly have that a=b=4. Comparing the coefficients of the x and units will give you two equations, in terms of c and d, which you can solve.

 

[tiny-tim]

… got to be careful …

C(x + 5) + D(x - 5) = 4x

Cx + 5c + dx + 5d = 4x

erm … no … Cx + 5c + dx - 5d = 4x!

Try again!

 

[the7joker7]

C + D = 4

and

5c - 5d = 0, then?

So that gives...

c = 2 and d = 2?

I just had another guy claim it was c = 4 and d = 0 using...

"using synthetic division divide numerator by denominator

x^2-25)4x^3+4x^2-96x-100(4x
...4x^3+0x^2-100x
______________________
......4x^2+4x-100(4
......4x^2+0x-100
_______________________
......4x

so (4x^3+4x^2-96x-100) /(x^2 - 25) = 4x + 4 + 4x/(x^2-25)

so a = 4, b = 4 , c = 4 and d = 0"

Which one's right?

 

[tiny-tim]

so (4x^3+4x^2-96x-100) /(x^2 - 25) = 4x + 4 + 4x/(x^2-25)

so a = 4, b = 4 , c = 4 and d = 0"

Which one's right?

Hi joker!

Well, he's sort-of right, and he sort-of isn't!

The first line above is correct - but it's exactly what you had anyway!

It isn't reduced to the simplest fractions!

So the second line is just optimistically re-defining c and d to fit the result!

You've gone one step further, and split the last fraction into two simpler ones.

Yours is defintitely right1

 

[the7joker7]

Thanks.

Hmm...Since derivative of denominator (x^2-25) is numerator, 2x , the integral of 2x dx/(x^2-25) is ln(x^2-25)

=>4x^2/2 + 4x + 2ln(x^2 -25) + c

=>2x^2 + 4x + 2ln[(x+5)(x-5)] + c

2x^2 + 4x + 2ln(x+5) + 2ln(x-5) + c

Does that sound in order?

 

[tiny-tim]

Hi joker!

Yes that's fine!

btw, I think the point "another guy" was correctly making was that, once you'd got 4x/(x^2 - 25), there was no point in breaking it down any further, since you could instantly see what its integral was!

But you had to go on in this case only because the question specifically required it.

 

[deezy911]

If it helps any...

Once you get your result from long division (should be 4x+4-(4x/x^2-25)), multiply through by x^2-25 on both sides. This will leave you with 4x^3+4x^2-96x-100 = 4x(x^2-25)+4(x^2-25)-4x.

Now set x = 0 , so A = 4

Now set x = 5 , so C = -1

Now set x = 1 , so B = -2 ( don't forget to plug in A and C to get B)

So your A B and C should be 4,-1, and -2 respectively.

 

[Count Iblis]

You don't need to solve a single equation.

f(x) = (4 x^3+4 x^2-96 x -100)/(x^2-25)

Expand around singular points and find asymptotic behavior at infinity:

Singular term in expansion around x = -5:

1/(x+5) * Lim x--->-5 of (x+5)f(x) = 2/(x+5)

Singular term in expansion around x = 5:

1/(x-5) Lim x--->5 of (x-5)f(x) = 2/(x-5)

Expansion around infinity:

The singular terms in this expanson (i.e. the terms that go to infinity as we approach the point around which we expand) are the positive powers of x. We have:

1/(x^2 - 25) = 1/x^2 1/[1-(5/x)^2] = 1/x^2 [1+25/x^2 + ...]

f(x) = 4 x + 4 + terms that go to zero for x to infinity

The sum of all the singular terms of the three expansions is:

g(x) = 2/(x+5) + 2/(x-5) + 4 x + 4

Consider the difference f(x) - g(x). Since both f and g are rational functions, f(x) - g(x) is a rational function. But it doesn't have any singularities as g contains the singular terms of the expansion around all the singular points of f. So, f(x) - g(x) is actually a polynomial. At infinity, f(x) - g(x) must tend to zero, as g(x) contains the positive powers of x in the large x expansion of f(x). This then implies that
f(x) - g(x) is zero.