Course Notes for Calculus II

\begin{align*} \frac{4}{(x-5)(x+3)} \amp = \frac{a}{x-5} + \frac{b}{(x+3)} \\ \amp = \frac{a(x+3) + b(x-5)}{(x-5)(x+3)} = \frac{(a+b)x + (3a-5b)}{(x-5)(x+3)} \end{align*}

\begin{align*} a+b \amp = 0 \\ 3a-5b \amp = 4 \\ a+b \amp = 0 \implies a = -b \\ 3a - 5b \amp = (3(-b) - 5b) = -8b = 4 \implies b = \frac{-1}{2}\\ a \amp = -b \implies a = \frac{1}{2} \end{align*}

\begin{align*} \frac{4}{(x-5)(x+3)} \amp = \frac{\frac{1}{2}}{x-5} + \frac{\frac{-1}{2}}{(x+3)} \\ \int \frac{4}{(x-5)(x+3)} dx \amp = \int \frac{\frac{1}{2}}{x-5} dx + \int \frac{\frac{-1}{2}}{(x+3)} dx \\ \amp = \frac{1}{2} \int \frac{1}{x-5} dx - \frac{1}{2} \int \frac{1}{x+3} dx \\ \amp = \frac{1}{2} \ln |x-5| - \frac{1}{2} \ln |x+3| + c \end{align*}

\begin{align*} \frac{3x^3+x-7}{(x-1)(x+3)} \amp = 3x - 6 + \frac{22x-25}{(x-1)(x+3)} \end{align*}

\begin{align*} \frac{22x-25}{(x-1)(x+3)} \amp = \frac{a}{x-1} + \frac{b}{x+3} = \frac{a(x+3) + b(x-1)}{(x-1)(x+3)}\\ \amp = \frac{ax + 3a + bx - b}{(x-1)(x+3)} = \frac{(a+b)x + (3a-b)}{(x-1)(x+3)} \end{align*}

\begin{align*} 22 \amp = a+b \implies a = 22-b \\ -25 \amp = 3a-b = 3(22-b) - b = -3b + 66 + b = 66 - 4b \\ -91 \amp = -4b \implies b = \frac{91}{4} \\ a \amp = 22-b = 22-\frac{91}{4} = \frac{88-91}{4} = \frac{-3}{4} \end{align*}

\begin{align*} \frac{22x-25}{(x-1)(x+3)} \amp = \frac{\frac{-3}{4}}{x-1} + \frac{\frac{91}{4}}{x+3} \\ \int \frac{3x^3 + x -7}{(x-1)(x+3)} dx \amp = \int 3x dx + \int 6 dx + \frac{-3}{4} \int \frac{1}{x-1} dx + \frac{91}{4} \int \frac{1}{x+3} dx \\ \amp = \frac{3x^2}{2} + 6x - \frac{3}{4} \ln |x-1| + \frac{91}{4} \ln |x+3| + c \end{align*}

\begin{align*} \frac{x^5 + 8x^3 - 2x - 1}{(x-4)(x+2)} dx \amp = x^3 + 2x^2 + 20 x + 56 + \frac{270x + 447}{(x-4)(x+2)} \end{align*}

\begin{align*} \frac{270x + 447}{(x-4)(x+2)} \amp = \frac{a}{x-4} + \frac{b}{x+2} = \frac{a(x+2) + b(x-4)}{(x+2)(x-4)} \\ \amp = \frac{ax + 2a + bx - 4b}{(x+2)(x-4)} = \frac{(a+b)x + (2a-4b)}{(x+2)(x-4)} \end{align*}

\begin{align*} 270 \amp = a+b \implies a = 270-b \\ 447 \amp = 2a - 4b = 2(270-b) - 4b = 540 -2b - 4b = 540 - 6b \\ -93 \amp = -6b \implies b = \frac{93}{6} = \frac{31}{2} \\ a \amp = 270-b = 270 - \frac{31}{2} = \frac{540 - 31}{2} = \frac{509}{2} \end{align*}

\begin{align*} \amp \frac{270x + 447}{(x-4)(x+2)} = \frac{\frac{509}{2}}{x-4} + \frac{\frac{31}{2}}{x+2} \\ \amp \int \frac{x^5 + 8x^3 - 2x - 1}{(x-4)(x+2)} dx \\ \amp = \int x^3 dx + \int 2x^2 dx + \int 20x dx + \int 56 dx + \frac{509}{2} \int \frac{1}{x-4} dx + \frac{31}{2} \int \frac{1}{x+2} dx\\ \amp = \frac{x^4}{4} + \frac{2x^3}{3} + 10x^2 + 56 x + \frac{509}{2} \ln |x-4| + \frac{31}{2} \ln |x+2| + c \end{align*}

\begin{align*} \frac{x^2}{(x-3)^3} \amp = \frac{a}{x-2} + \frac{b}{(x-2)^2} + \frac{c}{(x-2)^3} \\ \amp = \frac{a (x^2 -4x + 4) + b(x-2) + c}{(x-2)^3} \\ \amp = \frac{(a)x^2 + (-4a + b)x + (4a -2b + c)}{(x-2)^3} \end{align*}

\begin{align*} 1 \amp = a \\ (-4a + b) \amp = 0 \implies b = 4a = 4(1) = 4 \\ (4a - 2b + c) \amp = 0 \implies c = 2b-4a = 8 - 4 = 4 \end{align*}

\begin{align*} \frac{x^2}{(x-3)^3} \amp = \frac{1}{x-2} + \frac{4}{(x-2)^2} + \frac{4}{(x-2)^3} \\ \int \frac{x^2}{(x-2)^3} dx \amp = \int \frac{1}{x-2} dx + 4 \int \frac{1}{(x-2)^2} dx + 4 \int \frac{1}{(x-2)^3} dx \\ \amp = \ln |x-2| - 4 \frac{1}{x-2} - 2 \frac{1}{(x-2)^2} + c \\ \amp = \ln |x-2| - \frac{4}{x-2} - \frac{2}{(x-2)^2} + c \end{align*}

\begin{align*} \frac{2x^2+3x+4}{(x-1)(x-2)(x-3)} \amp = \frac{a}{x-1} + \frac{b}{x-2} + \frac{c}{x-3} \\ \amp = \frac{a(x-2)(x-3) + b(x-1)(x-3) + c(x-1)(x-2)}{(x-1)(x-2)(x-3)} \\ \amp = \frac{ax^2 - 5ax + 6a + bx^2 - 4bx + 3b + cx^2 - 3cx + 2c}{(x-1)(x-2)(x-3)} \\ \amp = \frac{(a+b+c)x^2 + (-5-4b-3c) x + (6a + 3b + 2c)}{(x-1)(x-2)(x-3)} \end{align*}

\begin{align*} 2 \amp = a + b + c \implies a = 2 - b - c \\ 3 \amp = (-5a-4b-3c) = -5(2-b-c) - 4b - 3c \\ \amp = -10 + 5b + 5c -4b - 3c = -10 + b + 2c \\ b \amp = 13 - 2c \\ a \amp - 2 - b - c = 2 - 13 + 2c - c = -11 + c \\ 4 \amp = 6a+3b+2c = 6(-11+c) + 3(13-2c) + 2c \\ \amp = -66 + 6c + 39 - 6c + 2c = -27 + 2c \\ 2c \amp = 31 \implies c = \frac{31}{2} \\ b \amp = 13 - 2c = 13 - 31 = -18 \\ a \amp = -11 + c = -11 + \frac{31}{2} = \frac{-22 + 31}{2} = \frac{9}{2} \end{align*}

\begin{align*} \amp \frac{2x^2+3x+4}{(x-1)(x-2)(x-3)} = \frac{\frac{9}{2}}{x-1} + -18 \frac{1}{x-2} dx + \frac{\frac{31}{2}}{x-3} \\ \amp \int \frac{2x^2 + 3x + 4}{(x-1)(x-2)(x-3)} dx = \frac{9}{2} \int \frac{1}{x-1} dx - 18 \int \frac{1}{x-2} dx + \frac{32}{2} \int \frac{1}{x-3} dx \\ \amp = \frac{9}{2} \ln |x-1| - 18 \ln |x-2| + \frac{31}{2} \ln |x-3| + c \end{align*}

\begin{align*} \amp \int \frac{1}{(x-2)^2(x-3)^2} = \frac{a}{x-2} + \frac{b}{(x-2)^2} + \frac{c}{(x-3)} + \frac{d}{(x-3)^2} \\ \amp = \frac{a(x-2)(x^2-6x+9) + b(x^2-6x+9) + c(x-3)(x^2-4x+4) + d(x^2-4x+4)}{(x-2)^2(x-3)^2} \\ \amp = \frac{a(x^3-8x^2+21x-18) + b(x^2-6x+9) + c(x^3-7x^2+16x-12) + d(x^2-4x+4)}{(x-2)^2(x-3)^2} \\ \amp = \frac{(a+c)x^3 + (-8a+b-7c+d)x^2 + (21a-6b+16c-4d)x + (-18a + 9b - 12c + 4d)}{(x-2)^2(x-3)^2} \end{align*}

\begin{align*} a + c \amp = 0 \implies a = -c \\ -8a+ b-7c+d \amp = 0 \\ c + b + d \amp = 0 \implies b = -d - c \\ 21a - 6b + 16c - 4d \amp = 0 \\ -5c - 6(-d-c) - 4d \amp = 0 \\ c + 2d \amp = 0 \implies c = -2d \\ -18a + 9b - 12c + 4d \amp = 1 \\ 6c + 9(-d-c) + 4d \amp = 1 \\ -3c - 5d \amp = 1\\ 6d - 5d \amp = 1 \implies d=1 \\ c \amp = -2d = -2 \\ b \amp = -d-c = 1 \\ a \amp = -c-= 2 \end{align*}

\begin{align*} \amp \frac{1}{(x-2)^2(x-3)^2} = \frac{2}{x-2} - \frac{1}{(x-2)^2} - \frac{2}{(x-3)} + \frac{d}{(x-3)^2} \\ \amp \int \frac{1}{(x-2)^2(x-3)^2} \\ \amp = 2 \int \frac{1}{x-2} dx + \int \frac{1}{(x-2)^2} dx - 2 \int \frac{1}{x-3} dx + \int \frac{1}{(x-3)^2}dx \\ \amp = 2 \ln |x-2| - \frac{1}{x-2} - 2 \ln |x-3| - \frac{1}{x-3} + c \end{align*}

\begin{align*} \amp \frac{2x^2+1}{(x-3)(x^2+x+1)} \\ \amp = \frac{a}{x-3} + \frac{bx+c}{x^2+x+1} = \frac{a(x^2+x+1) + (bx+c)(x-3)}{(x-3)(x^2+x+1)} \\ \amp = \frac{ax^2 + ax + a + bx^2 + cx - 3bx - 3c}{(x-3)(x^2+2+1)}\\ \amp = \frac{(a+b)x^2 + (a-3b+c)x + (a - 3c)}{(x-3)(x^2+x+1)} \end{align*}

\begin{align*} 2 \amp = a+b \implies a = 2-b \\ 0 \amp = a-3b+c \\ 0 \amp = (2-b) - 3b + c \\ 0 \amp = 2 -4b + c \implies c = 4b-2 \\ 1 \amp = a-3c \\ 1 \amp = (2-b) - 3(4b-2) \\ 1 \amp = 2 - b - 12 b + 6 \\ -7 \amp = -13 b \implies b= \frac{7}{13} \\ a \amp = 2 - b = 2 - \frac{7}{13} = \frac{19}{13} \\ c \amp = 4b - 2 = 4 \frac{7}{13} - 2 = \frac{28-26}{13} = \frac{2}{13} \end{align*}

\begin{align*} \frac{2x^2+1}{(x-3)(x^2+x+1)} \amp = \frac{1}{13} \left[ \frac{19}{x-3} + \frac{7x+2}{x^2+x+1} \right] \\ \int \frac{2x^2+1}{(x-3)(x^2+x+1)} dx \amp = \frac{1}{13} \left[ \int \frac{19}{x-3} dx + \int \frac{7x+2}{x^2+x+1} dx \right] \end{align*}

\begin{align*} \amp = \frac{19}{13} \ln |x-3| + \frac{1}{13} \int \frac{\frac{7}{2} \left(2x + \frac{4}{7} \right)}{x^2+ x+1} dx \\ \amp = \frac{19}{13} \ln |x-3| + \frac{7}{26} \int \frac{2x + 1}{x^2+ x+1} dx + \frac{7}{26} \int \frac{\frac{4}{7} - 1}{x^2 + x + 1} dx \\ \amp = \frac{19}{13} \ln |x-3| + \frac{7}{26} \int \frac{2x + 1}{x^2+ x+1} dx + \frac{7}{26} \int \frac{\frac{-3}{7}}{x^2+x+1} dx \\ \amp = \frac{19}{13} \ln |x-3| + \frac{7}{26} \ln |x^2+x+1| - \frac{3}{26} \int \frac{1}{ \left(x + \frac{1}{2} \right)^2 + \frac{3}{4}} dx \\ \amp = \frac{19}{13} \ln |x-3| + \frac{7}{26} \ln |x^2+x+1| - \frac{3}{26} \left( \frac{2}{\sqrt{3}} \arctan \left( \frac{2 \left( x + \frac{1}{2} \right)}{\sqrt{3}} \right) \right) + c \\ \amp = \frac{19}{13} \ln |x-3| + \frac{7}{26} \ln |x^2+x+1| - \frac{\sqrt{3}}{13} \arctan \left( \frac{2x+1}{\sqrt{3}} \right) + c \end{align*}

\begin{align*} \int \frac{x^4-x^2-1}{x(x^2+4x+5)} \frac{x^4-x^2-1}{x(x^2+4x+5)} \amp = x - 4 + \frac{10x^2 + 20x - 1}{x(x^2+4x+5)} \end{align*}

\begin{align*} \frac{10x^2 + 20x - 1}{x(x^2+4x+5)} \amp = \frac{a}{x} + \frac{bx+c}{x^2+4x+5} = \frac{a(x^2+4x+5) + (bx+c)(x)}{x(x^2+4x+5)} \\ \amp = \frac{ax^2 + 4ax + 5a + bx^2 + cx}{x(x^2+4x+5)} = \frac{(a+b)x^2 + (4a+c)x + 5a}{x(x^2+4x+5)} \end{align*}

\begin{align*} 5a \amp = -1 \implies a = \frac{-1}{5} \\ 4a + c \amp = 20 \implies c = 20 - 4a = 20 + \frac{4}{5} = \frac{104}{5}\\ a+b \amp = 10 \implies b = 10 - a = 10 - \frac{-1}{5} = \frac{51}{5} \end{align*}

\begin{align*} \frac{10x^2 + 20x}{x(x^2+4x+5)} \amp = \frac{1}{5} \left[ \frac{-1}{x} + \frac{51x+104}{x^2+4x+5} \right] \\ \int \frac{x^4-x^2-1}{x(x^2+4x+5)} \amp = \frac{-1}{5} \int \frac{1}{x} dx + \frac{1}{5} \int \frac{51x + 104}{x^2+4x+5} dx \end{align*}

\begin{align*} \amp = \frac{-1}{5} \ln |x| + \frac{1}{5}\frac{51}{2} \int \frac{2x + \frac{208}{51}}{x^2+4x+5} dx \\ \amp = \frac{-1}{5} \ln |x| + \frac{51}{10} \int \frac{2x + 4 + \frac{4}{51}}{x^2 + 4x + 5} dx \\ \amp = \frac{-1}{5} \ln |x| + \frac{51}{10} \int \frac{2x + 4}{x^2 + 4x + 5}dx + \frac{51}{10} \frac{4}{51} \int \frac{1}{x^2 + 4x +5} dx \\ \amp = \frac{-1}{5} \ln |x| + \frac{51}{10} \ln |x^2 + 4x + 5| + \frac{2}{5} \int \frac{1}{x^2 + 4x + 4 + 1} dx \\ \amp = \frac{-1}{5} \ln |x| + \frac{51}{10} \ln |x^2 + 4x + 5| + \frac{2}{5} \int \frac{1}{(x+2)^2 + 1} dx \\ \amp = \frac{-1}{5} \ln |x| + \frac{51}{10} \ln |x^2 + 4x + 5| + \frac{2}{5} \arctan (x+2) + c \end{align*}

\begin{equation*} \frac{x^2}{2} - 4x + \frac{-1}{5} \ln |x| + \frac{51}{10} \ln |x^2 + 4x + 5| + \frac{2}{5} \arctan (x+2) + c \end{equation*}

\begin{align*} \amp \frac{3}{(x-2)^2(x^2-x+3)} = \frac{a}{x-2} + \frac{b}{(x-2)^2} + \frac{cx+d}{x^2-x+3} \\ \amp = \frac{a(x-2)(x^2-x+3) + b(x^2-x+3) + (cx+d)(x-2)^2}{(x-2)^2(x^2-x+3)} \\ \amp = \frac{ax^3 - 3ax^2+5ax - 6a + bx^2 -bx + 3b + cx^3 + dx^2 - 4cx^2 - 4dx + 4cx + 4d}{(x-2)^2(x^2-x-3)}\\ \amp = \frac{(a + c)x^3 + (-3a + b - 4c + d) x^2 + (5a - b + 4c - 4d) x + (-6a + 3b + 4d)}{(x-2)^2 + (x^2 - x- 3)} \end{align*}

\begin{align*} a+c \amp = 0 \\ -3a + b - 4c + d \amp = 0 \\ 5a - b + 4c - 4d \amp = 0 \\ -6a + 3b + 4d \amp = 3 \end{align*}

\begin{align*} a \amp = \frac{-9}{25} \amp \amp b = \frac{3}{5} \amp \amp c = \frac{9}{25} \amp \amp d = \frac{-6}{25} \end{align*}

\begin{align*} \amp \frac{3}{(x-2)^2(x^2-x+3)} = \frac{-9}{25} \frac{1}{x-2} + \frac{3}{5} \frac{1}{(x-2)^2} + \frac{1}{25} \frac{9x-6}{x^2-x+3} \\ \amp \int \frac{3}{(x-2)^2(x^2-x+3)} dx \\ \amp = \frac{-9}{25} \int \frac{1}{x-2} dx + \frac{3}{5} \int \frac{1}{(x-2)^2} dx + \frac{1}{25} \int \frac{9x-6}{x^2-x+3} dx \end{align*}

\begin{align*} \amp = \frac{-9}{25} \ln |x-2| - \frac{3}{5} \frac{1}{x-2} + \frac{9}{50} \int \frac{2x -\frac{12}{9}}{x^2-x+3} dx \\ \amp = \frac{-9}{25} \ln |x-2| - \frac{3}{5} \frac{1}{x-2} + \frac{9}{50} \int \frac{2x - 1}{x^2-x+3} dx - \frac{9}{50} \frac{1}{3}\int \frac{1}{x^2-x+3} dx \\ \amp = \frac{-9}{25} \ln |x-2| - \frac{3}{5} \frac{1}{x-2} + \frac{9}{50} \ln |x^2-x+3| - \frac{3}{50} \int \frac{1}{\left( x-\frac{1}{2} \right)^2 + \frac{11}{4}} dx \\ \amp = \frac{-9}{25} \ln |x-2| - \frac{3}{5} \frac{1}{x-2} + \frac{9}{50} \ln |x^2-x+3| - \frac{3}{50} \frac{2}{\sqrt{11}} \arctan \left( \frac{ x - \frac{1}{2}}{\frac{\sqrt{11}}{2}} \right) + c \\ \amp = \frac{-9}{25} \ln |x-2| - \frac{3}{5} \frac{1}{x-2} + \frac{9}{50} \ln |x^2-x+3| - \frac{3}{25\sqrt{11}} \arctan \left( \frac{ 2x - 1}{\sqrt{11}} \right) + c \end{align*}

\begin{align*} \amp \frac{2x^2+4x-1}{(x^2+3x+4)(x^2+6)} = \frac{ax + b}{x^2+3x+4} + \frac{cx+d}{x^2+6} \\ \amp = \frac{(ax+b)(x^2+6) + (cx+d)(x^2+3x+4)}{(x^2+3x+4)(x^2+6)}\\ \amp = \frac{ax^3 + bx^2 + 6ax + 6b + cx^3 + dx^2 + 3cx^2 + 3dx + 4cx + 4d}{(x^2+3x+4)(x^2+6)} \\ \amp = \frac{(a+c)x^3 + (b + d + 3c)x^2 + (6a + 4c + 3d) x + (6b + 4d)}{(x^2+3x+4)(x^2+6)} \end{align*}

\begin{align*} a+c \amp = 0 \\ b +d + 3c \amp = 2 \\ 6a + 4c + 3d \amp = 4 \\ 6b + 4d \amp = -1 \end{align*}

\begin{align*} a \amp = \frac{-31}{58} \amp \amp b = \frac{-75}{58} \amp \amp c = \frac{31}{58} \amp \amp d = \frac{98}{58} \end{align*}

\begin{align*} \amp \frac{x^2+4x-1}{(x^2+3x+4)(x^2+6)} \frac{1}{58} \left[ \frac{-31x - 75}{x^2+3x+4} + \frac{31x+98}{x^2+6} \right] \\ \int \frac{-31x - 75}{x^2+3x+4} dx \amp = \frac{-31}{2} \int \frac{ 2x + \frac{150}{31}}{x^2+3x+4} dx \\ \amp = \frac{-31}{2} \int \frac{2x + 3 + \frac{57}{31}}{x^2 +3 x + 4} dx\\ \amp = \frac{-31}{2} \int \frac{2x+ 3}{x^2 +3 x +4} dx + \frac{-31}{2} \frac{57}{31} \int \frac{1}{x^2+3x+4}dx \\ \amp = \frac{-31}{2} \ln |x^2+3x+4| - \frac{57}{2} \int \frac{1}{x^2 + 3x + \frac{9}{4} + \frac{7}{4}} dx \\ \amp = \frac{-31}{2} \ln |x^2+3x+4| - \frac{57}{2} \int \frac{1}{ \left(x+\frac{3}{2} \right)^2 + \frac{7}{4}} dx \\ \amp = \frac{-31}{2} \ln |x^2+3x+4| - \frac{57}{2} \frac{2}{\sqrt{7}} \arctan \left( \frac{x + \frac{3}{2}}{\frac{\sqrt{7}}{2}} \right) + c \\ \amp = \frac{-31}{2} \ln |x^2+3x+4| - \frac{57}{\sqrt{7}} \arctan \left( \frac{2x + 3}{\sqrt{7}} \right) + c \end{align*}

\begin{align*} \int \frac{31x+98}{x^2+6} dx \amp = \frac{31}{2} \int \frac{2x + \frac{196}{31}}{x^2+6} dx \\ \amp = \frac{31}{2} \int \frac{2x}{x^2+6} dx + \frac{31}{2} \frac{196}{31} \int \frac{1}{x^2 + 6} dx \\ \amp = \frac{31}{2} \ln |x^2+6| + 98 \frac{1}{\sqrt{6}} \arctan \left( \frac{x}{\sqrt{6}} \right) + c \\ \amp = \frac{31}{2} \ln |x^2+6| + \frac{98}{\sqrt{6}} \arctan \left( \frac{x}{\sqrt{6}} \right) + c \end{align*}

\begin{align*} \int \frac{2x^2+4x-1}{(x^2+3x+4)(x^2+6)} \amp = \frac{1}{58} \left[ \int \frac{-31x - 75}{x^2+3x+4} dx + \int \frac{31x+98}{x^2+6} dx \right] \\ \amp = \frac{1}{58} \left[ \frac{-31}{2} \ln |x^2+3x+4| - \frac{57}{\sqrt{7}} \arctan \left( \frac{2x + 3}{\sqrt{7}} \right) \right.\\ \amp \left. + \frac{31}{2} \ln |x^2+6| + \frac{98}{\sqrt{6}} \arctan \left( \frac{x}{\sqrt{6}} \right) \right] + c \\ \amp = \frac{-31}{116} \ln |x^2+3x+4| - \frac{57}{58\sqrt{7}} \arctan \left( \frac{2x + 3}{\sqrt{7}} \right) \\ \amp + \frac{31}{116} \ln |x^2+6| + \frac{49}{29\sqrt{6}} \arctan \left( \frac{x}{\sqrt{6}} \right) + c \end{align*}