\begin{align*}\int \frac{1}{x+1}&+\frac{1}{x+2} \ dx\\&=\ln|x+1|+\ln|x+2|+C\\&=\ln|(x+1)(x+2)|+C\end{align*}

Oppgave 1

Delbrøkoppspalting er ikke en integralregel, men en metode for å skrive om uttrykket slik at det blir enklere å integrere.

Husk at \begin{align}(\ln x)’&=\frac{1}{x}\\
\int \frac{1}{x} \ dx &=\ln x + C\end{align}

  • \begin{align*}\int \frac{1}{x^2-1} \ dx&=\int \frac{1}{2}\frac{1}{(x-1)}-\frac{1}{2}\frac{1}{(x-1)} \ dx\\\text{Delbrøkoppspalting : }\\
    \frac{1}{(x+1)(x-1)}&=\frac{A}{x+1}+\frac{B}{x-1}\\1&=A(x-1)+B(x+1)\\x=-1 \ : \ 1&=-2A \Rightarrow A=-\frac{1}{2}\\x=1 \ : \ 1&=2B \Rightarrow B=\frac{1}{2}\\-----&-----\\&=\frac{1}{2}\int \frac{1}{(x-1)}-\frac{1}{(x-1)} \ dx\\&=\frac{1}{2}(\ln(x-1)-\ln(x+1))+C\\&=\frac{1}{2}\ln\Big|\frac{x-1}{x+1}\Big|+C\end{align*}

  • \begin{align*}\int \frac{4}{x^2-x} \ dx&=\int \frac{4}{x-1}-\frac{4}{x} \ dx\\\text{Delbrøkoppspalting : }\\
    \frac{4}{x^2-x}&=\frac{A}{x}+\frac{B}{x-1}\\4&=A(x-1)+Bx\\x=0 \ : \ 4&=-A \Rightarrow A=-4\\x=1 \ : \ 4&=B \Rightarrow B=4\\-----&-----\\&=4(\ln(x-1)-\ln x)+C\\&=4\ln \Big(\frac{x-1}{x}\Big)+C\end{align*}

  • \begin{align*}\int \frac{1}{x^2-4} \ dx&=\int\frac{1}{(x+2)(x-2)} \ dx\\\text{Delbrøkoppspalting : }\\
    \frac{1}{(x+2)(x-2)}&=\frac{A}{x+2}+\frac{B}{x-2}\\1&=A(x-2)+B(x+2)\\x=-2 \ : \ 1&=-4A \Rightarrow A=-\frac{1}{4}\\x=2 \ : \ 1&=4B \Rightarrow B=\frac{1}{4}\\-----&-----\\\int\frac{1}{(x+2)(x-2)} \ dx&=\int \frac{\frac{1}{4}}{x-2}-\frac{\frac{1}{4}}{x+2} \ dx\\&=\frac{1}{4}(\ln|x-2|-\ln|x+2|)+C\\&=\frac{1}{4}\ln\Big(\frac{x-2}{x+2}\Big)+C\end{align*}

  • \begin{align*}\int \frac{1}{2x^2-18} \ dx&=\int \frac{1}{2(x+3)(x-3)} \ dx\\\text{Delbrøkoppspalting : }\\
    \frac{1}{(x+3)(x-3)}&=\frac{A}{(x-3)}+\frac{B}{(x+3)}\\1&=A(x+3)+b(x-3)\\x=-3 \ : \ 1&=-6B \Rightarrow B=-\frac{1}{6}\\x=3 \ : \ 1&=6A \Rightarrow A=\frac{1}{6}\\-----&-----\\&=\frac{1}{2}\int \frac{1}{6}\frac{1}{(x-3)}-\frac{1}{6}\frac{1}{(x+3)} \ dx\\&=\frac{1}{12}\Big(\ln|x-3|-\ln|x+3|\Big)+C\\&=\frac{1}{12}\ln\Big(\frac{x-3}{x+3}\Big)+C\end{align*}

  • \begin{align*}\int \frac{2}{x^2-9} \ dx&=\int \frac{\frac{1}{3}}{x-3}+\frac{\frac{1}{3}}{x+3} \\&=\frac{1}{3}\int \frac{1}{x-3}-\frac{1}{x+3} \\&=\frac{1}{3}(\ln|x-3|-\ln|x+3|)+C\\&=\frac{1}{3}\ln\Big|\frac{(x-3)}{(x+3)}\Big|\\\text{Delbrøkoppspalting : }\\
    \frac{2}{(x-3)(x+3)}&=\frac{A}{x-3}+\frac{B}{x+3}\\2&=A(x+3)+B(x-3)\\x=3 \ : \ 2&=6A \Rightarrow A=\frac{1}{3}\\x=-3 \ : \ 2&=-6B \Rightarrow B=-\frac{1}{3}\\\end{align*}

  • \begin{align*}\int \frac{1}{x^2-16} \ dx&=\int \frac{A}{x+4}+\frac{B}{x-4} \ dx\\\text{Delbrøkoppspalting : }\\
    \frac{1}{x^2-16}&=\frac{A}{x+4}+\frac{B}{x-4}\\1&=A(x-4)+B(x+4)\\x=-4 \ : \ 1&=A(-8) \Rightarrow A=-\frac{1}{8}\\x=4 \ : \ 1&=B\cdot 8 \Rightarrow B=\frac{1}{8}\\-----&-----\\&=\int \frac{-\frac{1}{8}}{x+4}+\frac{\frac{1}{8}}{x-4} \ dx\\&=\frac{1}{8}\int \frac{1}{x-4}-\frac{1}{x+4} \ dx\\&=\frac{1}{8}(\ln|x-4|-\ln|x+4|)+C\\&=\frac{1}{8}\ln|\frac{x-4}{x+4}|+C\end{align*}

Oppgave 2

  • \begin{align*}\int \frac{5x+1}{x^2+x-2} \ dx&=\int \frac{A}{x+2}+\frac{B}{x-1} \ dx\\\text{Delbrøkoppspalting : }\\
    \frac{5x+1}{x^2+x-2}&=\frac{A}{x+2}+\frac{B}{x-1}\\5x+1&=A(x-1)+B(x+2)\\x=-2 \ : \ -9&=-3A \Rightarrow A=3\\x=1 \ : \ 6&=3B \Rightarrow B=2\\-----&-----\\&=\int \frac{3}{x+2}+\frac{2}{x-1} \ dx\\&=3\ln|x+2|+2\ln|x-1|+C\end{align*}

  • \begin{align*}\int \frac{3x+1}{x^2-x-6} \ dx&=\frac{3x+1}{(x-3)(x+2)} \ dx\\\text{Delbrøkoppspalting : }\\
    \frac{3x+1}{(x-3)(x+2)}&=\frac{A}{x-3}+\frac{B}{x+2}\\3x+1&=A(x+2)+B(x-3)\\x=3 \ : \ 3\cdot 3+1&=A(3+2)\Rightarrow 10=5A \Rightarrow A=2\\x=-2 \ : \ 3\cdot (-2)+1&=B(-2-3)\Rightarrow -5=-5B\Rightarrow B=1\\-----&-----\\\frac{3x+1}{(x-3)(x+2)} \ dx&=\int\frac{2}{x-3}+\frac{1}{x+2} \ dx\\&=2\ln|x-3|+\ln|x+2|+C\end{align*}

  • \begin{align*}\int \frac{x^2+x+13}{x^3-2x^2-5x+6} \ dx&=\\
    \text{Faktorisering av nevneren : }P(x)&=x^3-2x^2-5x+6\\P(1)&=1-2-5+6=0\\(x^3-2x^2-5x+6):(x-1)&=x^2-x-6=(x-3)(x+2)\\
    \text{Delbrøkoppspalting : }\\ \frac{x^2+x+13}{x^3-2x^2-5x+6}
    &=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{x-3}\\
    x^2+x+13=A(x+2)(x-3)+&B(x-1)(x-3)+C(x-1)(x+2)\\x=1 \\ : \ 15&=A\cdot 3\cdot (-2) \Rightarrow A=-\frac{15}{6}=-\frac{5}{2}\\x=-2 \\ : \ 15&=B\cdot (-3)\cdot (-5)\Rightarrow B=1\\x=3 \\ : \ 25&=C\cdot 2\cdot 5 \Rightarrow c=\frac{25}{10}=\frac{5}{2}\\-----&-----\\\int \frac{x^2+x+13}{x^3-2x^2-5x+6} \ dx&=\int -\frac{5}{2}\cdot \frac{1}{x-1}+\frac{1}{x+2}+\frac{5}{2}\cdot\frac{1}{x-3}\\ &=\ln|x+2|+\frac{5}{2}\ln|x-3|-\frac{5}{2}\ln|x-1|+C\end{align*}

Oppgave 3

  • \begin{align*}\int \frac{8}{x^3-4x} \ dx&=\int \frac{8}{x(x+2)(x-2)} \ dx\\\text{Delbrøkoppspalting : }\\
    \frac{8}{x(x+2)(x-2)}&=\frac{A}{x}+\frac{B}{x+2}+\frac{C}{x-2}\\8&=A(x+2)(x-2)+Bx(x-2)+Cx(x+2)\\x=0 \ : \ 8&=-4A \Rightarrow A=-2\\x=-2 \ : \ 8&=8B \Rightarrow B=1\\x=2 \ : \ 8&=8C \Rightarrow C=1\\-----&-----\\\int \frac{8}{x(x+2)(x-2)} \ dx&=\int -\frac{2}{x}-\frac{1}{x+2}+\frac{1}{x-2} \ dx\\&=\ln|x-2|+\ln|x+2|-2\ln|x|+C\\&=\ln|(x-2)(x+2)|-2\ln|x|+C\\&=\ln|(x-2)(x+2)|-\ln|x^2|+C\\&=\ln\Big|\frac{x^2-4}{x^2}\Big|+C\end{align*}

  • \begin{align*}\int \frac{x+4}{x^2+2x} \ dx&=\int \frac{2}{x}-\frac{1}{x+2}\\\text{Delbrøkoppspalting : }\\
    \frac{x+4}{x(x+2)}&=\frac{A}{x}+\frac{B}{x+2}\\x+4&=A(x+2)+B(x)\\x=0 \ : \ 4&=2A \Rightarrow A=2\\x=-2 \ : \ 2&= -2B \Rightarrow B=-1\\-----&-----\\&=2\ln|x|-\ln|x+2|+C\end{align*}

  • \begin{align*}\int \frac{3x}{x^2-x-2} \ dx&=\int \frac{A}{(x-2)}+\frac{B}{x+1} \ dx\\\text{Delbrøkoppspalting : }\\
    \frac{3x}{x^2-x-2}&=\frac{A}{(x-2)}+\frac{B}{x+1} \ dx\\3x&=A(x+1)+B(x-2)\\x=2 \ : \ 6&=3A \Rightarrow A=2\\x=-1 \ : \ -3&=-3B \Rightarrow B=1\\-----&-----\\&=\int \frac{2}{(x-2)}+\frac{1}{x+1} \ dx\\&=2\ln|x-2|+\ln|x+1|+C\end{align*}

Oppgave 4

Oppgave 5

  • \begin{align*}\int \frac{2x+4}{x^2+4x+3} \ dx&=\int \frac{A}{x+3}+\frac{B}{x+1} \ dx\\\text{Delbrøkoppspalting : }\\
    \frac{2x+4}{x^2+4x+3}&=\frac{A}{x+3}+\frac{B}{x+1}\\2x+4&=A(x+1)+B(x+3)\\x=-3 \ : \ -2&=-2A \Rightarrow A=1\\x=-1 \ : \ 2&=2B \Rightarrow B=1\\-----&-----\\&=\int \frac{1}{x+3}+\frac{1}{x+1} \ dx\\&=\ln|x+3|+\ln|x+1|+C\end{align*}

  • \begin{align*}\int \frac{3x-5}{x^2-x-12} \ dx&=\int \frac{A}{x-4}+\frac{B}{x+3} \ dx\\\text{Delbrøkoppspalting : }\\
    \frac{3x-5}{x^2-x-12}&=\frac{A}{x-4}+\frac{B}{x+3}\\3x-5&=A(x+3)+B(x-4)\\x=4 \ : \ 7&=7A \Rightarrow A=1\\x=-3 \ : \ -14&=-7B \Rightarrow B=2\\-----&-----\\&=\int \frac{1}{x-4}+\frac{2}{x+3} \ dx\\&=\ln|x-4|+2\ln|x+3|+C\end{align*}

  • \begin{align*}\int \frac{x^2+x-1}{x^2-x} \ dx&=\\\text{Polynomdivisjon : }\\(x^2+x-1):(x^2-x)&=1+\frac{2x-1}{x(x-1)}\\\text{Mellomregning : }\\
    \int \frac{2x-1}{x(x-1)} \ dx&=\int\frac{A}{x}+\frac{B}{x-1} \ dx\\\text{Delbrøkoppspalting : }\\ \frac{2x-1}{x(x-1)}&=\frac{A}{x}+\frac{B}{x-1}\\2x-1&=A(x-1)+Bx\\x=0 \ : \ -1&=-A \Rightarrow A=1\\x=1 \ : \ 1&=B \Rightarrow B=1\\-----&-----\\&=\int1+\frac{1}{x}+\frac{1}{x-1} \ dx\\&=x+\ln|x|+\ln|x-1|+C\end{align*}

Oppgave 6

  • \begin{align*}\int \frac{2x^2+5x+1}{x^2+x} \ dx&=\\\text{Polynomdivisjon : }(2x^2+5x+1)(x^2+x)&=2+\frac{3x+1}{x^2+x}\\\text{Mellomregning : }\int \frac{3x+1}{x(x+1)} \ dx&=\int \frac{A}{x}+\frac{B}{x+1} \ dx\\\text{Delbrøkoppspalting : }\\
    \frac{3x+1}{x(x+1)}&=\frac{A}{x}+\frac{B}{x+1}\\3x+1&=A(x+1)+Bx\\x=0 \ : \ 1&=A \Rightarrow A=1\\x=-1 \ : \ -2&=-B \Rightarrow B=2\\-----&-----\\&=\int \frac{1}{x}+\frac{2}{x+1}\\&=\ln|x|+2\ln|x+1|+C\\-----&-----\\&=\int 2+\frac{3x+1}{x^2+x} \ dx\\&=2x+\ln|x|+2\ln|x+1|+C\end{align*}

  • \begin{align*}\int \frac{2x^3+x^2-2x-3}{x^2-1} \ dx&=\int \frac{2x^3+x^2-2x-3}{(x+1)(x-1)} \ dx\\\text{Polynomdivisjon : }(2x^3+x^2-2x-3):(x^2-1)&=2x+1-\frac{2}{x^2-1}\\-----&-----\\\text{Mellomregning : }\int \frac{2}{x^2-1} \ dx&=\int \frac{A}{x-1}+\frac{B}{x+1} \ dx\\&=\int \frac{1}{x-1}-\frac{1}{x+1} \ dx\\&=\ln|x-1|-\ln|x+1|+C\\\text{Delbrøkoppspalting : }\frac{2}{(x+1)(x-1)}&=\frac{A}{x-1}+\frac{B}{x+1}\\2&=A(x+1)+B(x-1)\\x=1 \ : \ 2&=2A \Rightarrow A=1\\x=-1 \ ; \ 2&=-2B \Rightarrow B=-1\\-----&-----\\\int 2x+1-\frac{2}{x^2-1} \ dx &=\int 2x+1 \ dx -2\int \frac{2}{x^2-1} \ dx\\&=x^2+x-2(\ln|x-1|-\ln|x+1|)+C\\&= x^2+x+2\ln|x+1|-2\ln|x-1|+C\\&=x^2+x+2\ln|\frac{x+1}{x-1}|+C\end{align*}