5.9 Integrals of a linear function divided by a quadratic

We now study the integral I =\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits (px + q)∕({x}^{2} + ax + b)\kern 1.66702pt dx, i.e., linear over quadratic, where the quadratic does not factorize.

• Calculate the differential of the denominator,

  {d\over dx}({x}^{2} + ax + b) = 2x + a.

  Use this to rearrange the numerator into form

  {p\over 2}(2x + a) + (q − pa∕2),

  i.e., as a constant times the derivative of the denominator plus another constant. We can now split the integral,

  I = {p\over 2}\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits {2x + a\over { x}^{2} + ax + b}\kern 1.66702pt dx + (q − pa∕2)\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits {dx\over { x}^{2} + ax + b}\quad .

  The first integral on the r.h.s. can be done using the substitution z = {x}^{2} + ax + b,

  \mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits {2x + a\over { x}^{2} + ax + b}\kern 1.66702pt dx =\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits {1\over z}\kern 1.66702pt dz =\mathop{ ln}\nolimits z =\mathop{ ln}\nolimits ({x}^{2} + ax + b)\quad .

• The second integral is more complicated, and we deal with

  J =\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits {dx\over { x}^{2} + ax + b}

  separately. The technique used is based on completing the square, {x}^{2} + ax + b ={ (x + c)}^{2} ± {d}^{2}, which after the substitution z = x + c leads to a standard integral

  \mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits {dz\over { z}^{2} ± {d}^{2}}.

  Depending on the sign we get either an inverse tangent or a ratio of logarithms,

  \begin{eqnarray*} \mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits {1\over { z}^{2} + {d}^{2}}dz& =& {1\over d}{\mathop{tan}\nolimits }^{−1}(z∕d) + c\kern 1.66702pt , %& \\ \mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits {1\over { z}^{2} − {d}^{2}}dz& =& {1\over 2d}\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits \left ( {1\over z − d} − {1\over z + d}\right )dz = {1\over 2d}\mathop{ln}\nolimits \left ({z − d\over z + d}\right ) + c\kern 1.66702pt .%& \\ \end{eqnarray*}

Example 5.18: 

Evaluate I =\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits {4x − 1\over { x}^{2} + 2x + 3}\kern 1.66702pt dx.

Solution: 

• Differentiating the denominator gives 2x + 2. Take apart into tow pieces, by rearranging numerator as 4x − 1 = 2(2x + 2) − 5.
  \begin{eqnarray*} I& =& \mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits {2x + 2\over { x}^{2} + 2x + 3}\kern 1.66702pt dx − 5\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits {dx\over { x}^{2} + 2x + 3}%& \\ & =& \mathop{ln}\nolimits ({x}^{2} + 2x + 3) − 5\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits {dx\over { x}^{2} + 2x − 3} %& \\ \end{eqnarray*}

  Now complete the square for the denominator, and find that

  \begin{eqnarray*}{ x}^{2} + 2x + 3& =& {(x + 1)}^{2} + 2 = {(x + 1)}^{2} +{ \sqrt{2}}^{2}%& \\ \end{eqnarray*}

  J =\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits {dx\over { x}^{2} + 2x + 3} =\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits {dx\over { (x + 1)}^{2} +{ \sqrt{2}}^{2}}.

  Substitute z = x + 1, dz = (dz∕dx)dx = dx,

  \begin{eqnarray*} J& =& \mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits {dz\over { z}^{2} +{ \sqrt{2}}^{2}} %& \\ & =& (1∕\sqrt{2}){\mathop{tan}\nolimits }^{−1}(z∕\sqrt{2}) + c\quad .%& \\ \end{eqnarray*}

  Thus we find

  \class{boxed}{I = 2\mathop{ln}\nolimits ({x}^{2} + 2x + 3) − (5∕\sqrt{2}){\mathop{tan}\nolimits }^{−1}((x + 1)∕\sqrt{2}) + c\quad . }

5.9.1 Completing the Square

Completing the square is a simple idea that is surprisingly useful. First a definition:

Let us look at a few examples:

A polynomial of infinite degree is usually called an infinite power series.

Any polynomial of degree 2, i.e., a quadratic, can always be rearranged to have the form a{(x + b)}^{2} + c, as the square of a linear term plus a constant. Bringing a quadratic polynomial to this form is called completing the square.

5.9.2 Method

In general, if two polynomials are equal, it means that the coefficient of each power of the variable are equal, since each power varies at a different rate with the variable. So in order to complete the square, we must equate coefficients of powers of x on both sides. We shall do this by example.

1. Complete the square in {x}^{2} + x + 1:
   Put
   \begin{eqnarray*}{ x}^{2} + x + 1& =& a{(x + b)}^{2} + c %& \\ & =& a{x}^{2} + 2abx + c + a{b}^{2}.%& \\ \end{eqnarray*}

   Now equate coefficients of {x}^{2} on both sides. We find 1 = a, or a = 1. Then compare the coefficients of x. We conclude 1 = 2ab. Using a = 1 we find b = 1∕2. Now equate the constant term, 1 = a{b}^{2} + c = {1\over 4} + c. We conclude that c = 3∕4.
   Collecting all the results we find

   \class{boxed}{{x}^{2} + x + 1 ={ \left (x + {1\over 2}\right )}^{2} + {3\over 4}. }

2. Complete the square in 2{x}^{2} − x.
   Solve 2{x}^{2} − x = a{(x + b)}^{2} + c. We compare coefficients of
   

   {x}^{2}:	2 = a,
   x:	− 1 = 2ab,	therefore	b = −1∕4,
   const:	0 ={ ab}^{2} + c,	therefore	c = −1∕8.

   
   Thus

   \class{boxed}{2{x}^{2} − x = 2{\left (x −{1\over 4}\right )}^{2} −{1\over 8}\quad . }

It is often useful to write the constant as