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Question: $\int \dfrac{5 }{(x+1) (x^2 + 4) } dx $

Thought process: I'm treating it as a partial fraction since it certainly looks like one.

I cannot seem to solve it besides looking at it in the "partial fraction" way.

My work: 1) Focus on the fraction part first ignoring the $\int $ and $dx$ for the moment.

Multiply $(x+1) (x^2+4)$ on both sides of the equation and get:

$5$ = $\dfrac{A }{(x+1) } $ + $\dfrac{Bx+C }{(x^2+4) } $

Note: the x^2 + 4 is irreducible which explains the Bx+c as the numerator.

$ 5 = A(x^2+ 4) + Bx + C(x+1) $

I tried x = -1 which knocks out C:

$5 = A5 + - B $

I also tried x = 0 which knocks out B

$5 = A4 + C$

A is a lot harder to knock out since the squared changes the picked value to be positive.

I decided to add the two found equation together and get

$10 = A9 - B + C $

Now I am officially stuck at this step.

Nicholas
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    I think you meant to group the terms $(Bx+C)$ so that your work reads $5 = A(x^{2}+4) + (Bx+C)(x+1)$. But your work looks good to me otherwise. – Alex Wertheim May 21 '14 at 22:31
  • I'm having trouble figuring out a value of x that would knock all coefficients except one of them. Is there a good strategy? – Nicholas May 21 '14 at 22:40
  • That approach works best when you have only linear factors in the numerator, but it can still be applied. Use $ \ x \ = \ -1 \ $ to get at $ \ A \ $ , then continue with simple values like $ \ x \ = \ 0 \ $ and $ \ x \ = \ 1 \ $ to finish the job of solving for $ \ B \ $ and $ \ C \ $ . – colormegone May 21 '14 at 22:49
  • You actually can separate the $x^2 + 4$, but it will introduce imaginary numbers--which will be difficult to get rid of after you take the integral. – Jared May 21 '14 at 22:49
  • You shouldn't be choosing values of $x$ at all--that's not the correct way to do partial fraction decomposition--it's a trick they teach you that only works if you know exactly what you're doing (for instance if you set up the problem incorrectly you'll get a solution but upon inspection it will be incorrect). – Jared May 21 '14 at 22:51
  • The last equation is not correct. You should have $(x^2+4)(x+1)$ on the left and $(x+1)$ multiplying the $Bx$ – Ross Millikan May 21 '14 at 23:03
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    Don't change the value of x, expand the polynomials on each side and use the coefficients of the same power to form simultaneous equations. – Graham Kemp May 21 '14 at 23:22
  • See my answer to this question for a variant technique. – Anant May 22 '14 at 19:27

3 Answers3

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Split the fraction. $$\color{blue}{\dfrac{5 }{(x+1)(x^2 + 4)} = \dfrac{P(x)}{(x+1)}+\dfrac{Q(x)}{(x^2+4)}}$$

The largest partial fraction's denominator is an order 2 polynomial, so the order of the partial fractions' numerators must become order 2 when cross multiplied with the other partial fractions' denominators. Ie: P(X) is order 0, Q(X) is order 1.

$$\color{blue}{\dfrac{5 }{(x+1)(x^2 + 4)} = \dfrac{A}{(x+1)}+\dfrac{Bx+C}{(x^2+4)}}$$

Now that you know what the target looks like, perform the crossmultiplication.

$$\color{blue}{5 = A(x^2+4)+(Bx+C)(x+1)}$$

Expand the polynomial so we can compare the coefficients.

$$\color{blue}{5 = (A+B)x^2 + (B+C)x + (4A+C)}$$

Which allows us to conclude $$\color{blue}{A=-B, C=-B, 5= 4A+C \implies B=-1, A=1, C=1}$$

So we have our partial fraction expansion. $$\color{blue}{\therefore \dfrac{5 }{(x+1)(x^2 + 4)} = \dfrac{1}{(x+1)}+\dfrac{(x+1)}{(x^2+4)}}$$

Next use: $\color{blue}{\dfrac{\mathrm{d}}{\mathrm{d}x}\ln(x+a) = \dfrac{1}{x+a}}$, $\color{blue}{\dfrac{\mathrm{d}}{\mathrm{d}x}\ln(x^2+b^2)= 2\dfrac{x}{x^2+b^2}}$ and $\color{blue}{\dfrac{\mathrm{d}}{\mathrm{d}x} \arctan \dfrac{x}{c} = \dfrac{c}{x^2+c^2}}$

Graham Kemp
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Your equation is incorrect--you have some algebra mistakes that are causing you to not get a solution (which is obvious unless you start plugging in values to $x$):

$$ \frac{5}{(x + 1)(x^2 + 4)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 4} $$

Now cross multiply to get the numerator:

\begin{align} 5 =& A(x^2 + 4) + (Bx + C)(x + 1) = Ax^2 + 4A + Bx^2 + Bx + Cx + C \\ =& (A + B)x^2 + (B + C)x + (4A + C) \end{align}

This yields exactly three equations for three unkowns. You don't plug in values, you match coefficients: $5 = 0*x^2 + 0 * x + 5$ therefore you have:

\begin{align} (A + B)x^2 =& 0*x^2& &\longrightarrow &A + B = 0 \\ (B + C)x =& 0*x&& \longrightarrow &B + C = 0 \\ 4A + C =& 5 && \longrightarrow & 4A + C = 5 \\ \end{align}

Solving those equations yields:

\begin{align} A + B = 0 &\longrightarrow A = -B \\ B + C = 0 & \longrightarrow C = -B = A\\ 4A + C = 5 &\longrightarrow 4A + A = 5 \longrightarrow A = 1, B = -1, C = 1\\ \end{align}

Solve those equations and you will get the correct partial fraction decomposition--no need to plug in arbitrary values of $x$.

$$ \frac{5}{(x + 1)(x^2 + 4)} = \frac{1}{x + 1} + \frac{1 - x}{x^2 + 4} = \frac{1}{x + 1} + \frac{1}{x^2 + 4} - \frac{x}{x^2 + 4} $$

Jared
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Note as in the comments you will have

$$ \dfrac{5 }{(x+1) (x^2 + 4) } = \frac{A}{x+1} + \frac{Bx + C}{x^2+4} $$

Cross multiplying and solving the equations gives

$$ \dfrac{5 }{(x+1) (x^2 + 4) } = \frac{1}{x+1} - \frac{x - 1}{x^2+4} $$

Which is easier to integrate. A sneakier way than solving the equations is the following

$$ \frac{5}{(x+1)(x^2+4)} = \frac{(x^2 + 4) + (1 - x^2)}{(x+1)(x^2+4)} = \frac{1}{x+1} + \frac{(1-x)(1+x)}{(x+1)(x^2+4)} $$ As wanted.

N3buchadnezzar
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  • I don't see how that's "sneaky" unless you somehow know the solution a priori. I mean other than the fact that it works, how would you synthesize that numerator without solving the equations? – Jared May 22 '14 at 02:01
  • @Jared, The "sneaky" bit is noticing that $5=4+1$, which leads to $5=\color{blue}{(x^2+4)}+(1-x)\color{blue}{(1+x)}$. That is, relating the numerator to terms appearing in the denominator. It relies on anticipating that the partial fractions will be "neat"; the way textbook and exam questions often are. Spotting such patterns can save time during exams. – Graham Kemp May 22 '14 at 03:55
  • @GrahamKemp OK, I'll give you that--you have to admit it's sort of "cheating", but you are right about test taking skills. – Jared May 22 '14 at 04:33
  • I personally can spot these 10 out of 10 times. Simply because I have spent so long doing it. But yeah it does not take much practice and is often much faster than solving the simple equations. – N3buchadnezzar May 22 '14 at 08:09