5.9 Integrals of a linear function divided by a quadratic

We now study the integral I=∫(px+q)∕(x2+ax+b)dx, i.e., linear over quadratic, where the quadratic does not factorize.

• Calculate the differential of the denominator,

  ddx(x2+ax+b)=2x+a

  Use this to rearrange the numerator into form

  2p(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=2p∫2x+ax2+ax+bdx+(q−pa∕2)∫dxx2+ax+b

  The first integral on the r.h.s. can be done using the substitution z=x2+ax+b,

  ∫2x+ax2+ax+bdx=∫z1dz=lnz=ln(x2+ax+b)

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

  J=∫dxx2+ax+b

  separately. The technique used is based on completing the square, x2+ax+b=(x+c)2±d2, which after the substitution z=x+c leads to a standard integral

  ∫dzz2±d2

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

  ∫1z2+d2dz∫1z2−d2dz==d1tan−1(z∕d)+c12d∫1z−d−1z+ddz=12dlnz−dz+d+c 

Example 5.18: 

Evaluate I=∫4x−1x2+2x+3dx.

Solution: 

• Differentiating the denominator gives 2x+2. Take apart into tow pieces, by rearranging numerator as 4x−1=2(2x+2)−5.
  I==∫2x+2x2+2x+3dx−5∫dxx2+2x+3ln(x2+2x+3)−5∫dxx2+2x−3

  Now complete the square for the denominator, and find that

  x2+2x+3=(x+1)2+2=(x+1)2+22 

  J=∫dxx2+2x+3=∫dx(x+1)2+22

  Substitute z=x+1, dz=(dz∕dx)dx=dx,

  J==∫dzz2+22(1∕2)tan−1(z∕2)+c

  Thus we find

  I=2ln(x2+2x+3)−(5∕2)tan−1((x+1)∕2)+c

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:

polynomial	degree
(a) x+1	1	Also called linear, since if we plot
(b) 4x	1	the functions y=x+1, y=4x, etc.
(c) ax+b	1	we get a straight line
(d) x2+2x+1	2
(e) −7x2−3	2	(also known as quadratic)
(f) ax2+bx+c	2
(g) x3∕9−πx	3	cubic
(h) 12x6+0001	6

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 x2+x+1:
   Put
   x2+x+1==a(x+b)2+cax2+2abx+c+ab2

   Now equate coefficients of x2 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=ab2+c=41+c. We conclude that c=3∕4.
   Collecting all the results we find

   x2+x+1=x+212+43

2. Complete the square in 2x2−x.
   Solve 2x2−x=a(x+b)2+c. We compare coefficients of
   

   x2:	2=a,
   x:	−1=2ab,	therefore	b=−1∕4,
   const:	0=ab2+c,	therefore	c=−1∕8.

   
   Thus

   2x2−x=2x−412−81

It is often useful to write the constant as

c=d2−d2(if c is positive)(if c is negative)