Separate each of the following into the sum of a polynomial and a sum of partial fractions.

• [math]\frac{5}{(x-2)(x+3)}[/math]
• [math]\frac{x+2}{(2x+1)(x+1)}[/math]
• [math]\frac{2x^3+3x^2-2}{2x^2+3x+1}[/math]
• [math]\frac{4x^2-5x+10}{(x-4)(x^2+2)}[/math]
• [math]\frac{3x^3+5x^2-27x+8}{x^2+4x}[/math]
• [math]\frac{x^2+1}{x^2+x+1}[/math].

Integrate each of the following.

• [math]\int \frac{5}{(x-2)(x+3)} \; dx[/math]
• [math]\int \frac{x+2}{(2x+1)(x+1)} \; dx[/math]
• [math]\int \frac{2x^3+3x^2-2}{2x^2+3x+1} \; dx[/math]
• [math]\int \frac{4x^2-5x+10}{(x-4)(x^2+2)} \; dx[/math].

Find the partial fractions decomposition of each of the following rational functions.

• [math]\frac{x-8}{x^2-x-6}[/math]
• [math]\frac{18}{x^2+8x+7}[/math]
• [math]\frac{x+1}{(x-1)^2}[/math]
• [math]\frac{8x+25}{x^2+5x}[/math]
• [math]\frac{4}{x^2(x+2)}[/math]
• [math]\frac{6x^2-x+13}{(x+1)(x^2+4)}[/math]
• [math]\frac{(x+2)^2}{(x+3)^3}[/math]
• [math]\frac{x^2+2x+5}{(2x-1)(x^2+1)^2}[/math].

Evaluate each of the following integrals.

• [math]\int \frac{x-8}{x^2-x-6} \; dx[/math]
• [math]\int \frac{18}{x^2+8x+7} \; dx[/math]
• [math]\int \frac{x+1}{(x-1)^2} \; dx[/math]
• [math]\int \frac{8x+25}{x^2+5x} \; dx[/math]
• [math]\int \frac{6x^2-x+13}{(x+1)(x^2+4)} \; dx[/math]
• [math]\int \frac{(x+2)^2}{(x+3)^3} \; dx[/math].

• lab{7.4.5a} Show directly that [math]\frac{2x-3}{(x-2)^2}[/math] can be written in the form [math]\frac{A}{x-2} + \frac{B}{(x-2)^2}[/math] by first writing [math]\frac{2x-3}{(x-2)^2} = \frac{2(x-2)+1}{(x-2)^2}[/math].
• lab{7.4.5b} Following the method in \ref{ex7.4.5a}, show that [math]\frac{ax+b}{(x-k)^2}[/math] can always be written [math]\frac{A}{x-k} + \frac{B}{(x-k)^2}[/math], where [math]A[/math] and [math]B[/math] are constants.
• Extend the result in \ref{ex7.4.5b} by factoring, completing the square, and dividing to show directly that
  [[math]] \frac{ax^2+bx+c}{(x-k)^3} \: \mbox{can be written} \: \frac{A}{x-k}+\frac{B}{(x-k)^2}+\frac{C}{(x-k)^3} [[/math]]
  where [math]A[/math], [math]B[/math] and [math]C[/math] are constants.

Why can there not be an irreducible cubic polynomial with real coefficients?

Integrate each of the following.

• [math]\int \frac{(3x+1)\;dx}{x^3+2x^2+x}[/math]
• [math]\int \frac{(x^2+1)\;dx}{x^2-3x+2}[/math]
• [math]\int \frac{(x-2)\;dx}{(2x+1)(x^2+1)}[/math]
• [math]\int \frac{x^2-3x-2}{(x-2)^2(x-3)}\;dx[/math]
• [math]\int \frac{dx}{x^2+2x+2}[/math]
• [math]\int \frac{(2x+1)\;dx}{x^2+2x+2}[/math]
• [math]\int \frac{\sec^2x\;dx}{\tan^2x-4\tan x+3}[/math]
• [math]\int \frac{\sec y \tan y \; dy}{2\sec^2y+5\sec y+2}[/math]
• [math]\int \frac{y^2+1}{y^2+y+1}\;dy[/math]
• [math]\int \frac{10+5z-z^2}{(z+4)(z^2+z+1)}\;dz[/math]
• [math]\int \frac{(6x+3)\;dx}{(x-1)(x+2)(x^2+x+1)}[/math]
• [math]\int \frac{(x^3+4)\;dx}{(x+1)(x+2)^2}[/math]
• [math]\int \frac{x^2+2x+5}{(2x-1)(x^2+1)^2}\;dx[/math]
• [math]\int \frac{dx}{(x^2+x+5)^3}[/math].

Prove that the statement in the text, \secref{7.4} that, since a nonzero polynomial of degree [math]n[/math] has at most [math]n[/math] distinct roots, two rational functions with the same denominator are equal if and only if their numerators are equal. [Hint: Suppose that [math]\frac{N_1(x)}{D(x)} = \frac{N_2(x)}{D(x)}[/math], where polynomial [math]D(x)[/math] is not the zero function. Then [math]\frac{N_1(x)-N_2(x)}{D(x)}=0[/math], and so the polynomial equation [math]N_1(x) - N_2(x) = 0[/math] holds for every real number [math]x[/math] for which [math]D(x) \ne 0[/math].]