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I have been using the cover up method from this video lecture

$$\int \frac{x^2 + 11x}{(x-1)(x+1)^2} dx$$

$$\frac{x^2 + 11x }{(x-1)(x+1)^2} = \frac{A}{x+1} + \frac{B}{x-1} + \frac{C}{x-1}$$

What do I do with the double $x-1$ denominator? With the cover up method it seems useless. I can get that $A$ is $3$, $B$ might be 5 but then so is $C$? How do I use the cover up method to solve this?

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  • $\begingroup$ When the denominator has repeated roots you can't determine all the coefficients using the cover up method. Look at the example which is covered at around 19 min into the lecture, he explains it quite well. $\endgroup$
    – Vincent Pfenninger
    Jul 4, 2013 at 22:26

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In general, for any factor in the denominator $(ax + b)^n$, where $n > 1$, we need to account for each multiplicity of factors: $$\frac{A}{ax + b} + \frac{B}{(ax + b)^2} + \cdots + \frac Z{(ax + b)^n}$$

So...You need to decompose as follows:

$${x^2 + 11 x\over (x-1){\bf (x+1)^2}} = {A\over x - 1} + {B\over {\bf x + 1}} + {C\over {\bf (x+1)^2}}.$$

Then essentially we solve for $$A(x+1)^2 + B(x-1)(x+ 1) + C(x-1) = x^2 + 11x + 0$$ by matching up coefficients. (It's like how we find the greatest common denominator, which is then set equal to the numerator of the original integrand.

Simpler yet:

$x = 1 \implies 4A = 12 \implies A = 3.\;$

Putting $x = -1 \implies -2C = -10 \implies C = 5$.

Now just solve for B: $x = 0 \implies A - B - C = 0 \implies B = A-C = 3-5 = -2$.

This gives us $${x^2 + 11 x\over (x-1)(x+1)^2} = {3\over x - 1} - {2\over x + 1} + {5\over (x+1)^2}$$

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  • $\begingroup$ I am trying to learn how to do the cover up method, and I get stuck on how to solve for C. $\endgroup$
    – Sailor Jim
    Jul 4, 2013 at 22:26
  • $\begingroup$ I'm not entirely sure what the cover up method is. Do you see the changed denominators from what you posted? The denominators here are what you need to work with. Do you see why $$A(x+1)^2 + B(x-1)(x+ 1) + C(x-1) = x^2 + 11x + 0$$ and how putting $x = -1$ gives us $A\cdot 0 + B\cdot 0 -2C = 1 - 11 = -10$, so that $C = 5$? $\endgroup$
    – amWhy
    Jul 4, 2013 at 22:30
  • $\begingroup$ I'll be happy to clarify anything along the way, and answer more questions about this post. $\endgroup$
    – amWhy
    Jul 4, 2013 at 22:31
  • $\begingroup$ This is a slight variation on the cover-up method, which is based on the same logic as is the cover-up method. But here, we can have $3$ unknowns $A, B, C$, and can solve for each. First, I "zeroed" out the $x +1$ factors by putting $x = -1$ to solve for A. Then I zeroed out the $x - 1$ factors by putting $x = 1$. Then, putting x = 0 made calculating the middle unknown easy. $\endgroup$
    – amWhy
    Jul 4, 2013 at 22:40
  • $\begingroup$ @Sailor Jim, this is what you are doing when you use Heaviside's cover-up method, cover-up is just a disguised version of this $\endgroup$
    – Matthew Hampsey
    Jul 5, 2013 at 6:13
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The third denominator should be $(x+1)^2$. When you have repeated factors in the denominator, you need one term with each power. The numerator can still be a constant.

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    $\begingroup$ or possibly $(x+1)^2$ $\endgroup$
    – robjohn ♦
    Jul 4, 2013 at 22:19
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You need to decompose as follows $${x^2 + 11 x\over (x-1)(x+1)^2} = {A\over x + 1} + {B\over x - 1} + {C\over (x+1)^2}.$$

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