$$\int u’\cdot v \ dx = u\cdot v-\int u \cdot v’ \ dx$$

Oppgave 1

  • Løsningsforslag:\begin{align*}\int 2x\cdot \ln(x) \ dx&=x^2\ln x-\int x^2\cdot \frac{1}{x} \ dx\\\text{Delvis integral : }&u=\ln x \ , \ u'=\frac{1}{x} \\&v=x^2 \ , \ v'=2x \\-----&-----\\&=x^2\ln x-\int x \ dx\\&=x^2\ln x-\frac{1}{2}x^2+C\\&=\frac{1}{2}x^2(2\ln x-1)+C\end{align*}

  • Løsningsforslag:\begin{align*}\int x^4\cdot \ln(x) \ dx&=\frac{1}{5}x^5 \ln x-\frac{1}{5}\int x^5\cdot \frac{1}{x} \ dx \\\text{Delvis integral : }&u=\ln(x) \ , \ u'=\frac{1}{x} \\&v=\frac{1}{5}x^5 \ , \ v'=x^4 \\-----&-----\\&=\frac{1}{5}x^5 \ln x-\frac{1}{5}\int x^4 \ dx\\&=\frac{1}{5}x^5 \ln x-\frac{1}{5}\cdot \frac{1}{5}x^5 + C\\&=\frac{1}{5}x^5(\ln x-\frac{1}{5})+C\end{align*}

  • Løsningsforslag:\begin{align*}\int \ln(x) \ dx&=\int 1\cdot \ln x \ dx\\\text{Delvis integral : }&u=\ln x \ , \ u'= \frac{1}{x} \\&v= x \ , \ v'= 1 \\-----&-----\\&=x\ln x-\int x\cdot \frac{1}{x} \ dx\\&=x\ln x -\int 1 \ dx\\&=x\ln x -x+C\end{align*}

Oppgave 2

  • Løsningsforslag:\begin{align*}\int 2x\cdot e^x \ dx&=2x e^x-2\int e^x \ dx\\\text{Delvis integral : }&u=2x \ , \ u'=2 \\&v=e^x \ , \ v'=e^x \\-----&-----\\&=2x e^x-2e^x+C\\&=2e^x(x-1)+C\end{align*}

  • Løsningsforslag:\begin{align*}\int e^x(x^2-1) \ dx&=e^x(x^2-1)-\int 2x e^x \ dx\\\text{Delvis integral : }\\&u=x^2-1 \ , \ u'= 2x \\&v= e^x \ , \ v'=e^x \\-----&-----\\\text{Mellomregning : }\\
    \int 2x e^x \ dx&=2x e^x-2\int e^x \ dx\\&=2xe^x-2e^x+C\\-----&-----\\\end{align*}\begin{align*}&=e^x(x^2-1)-(2xe^x-2e^x)+C\\&=e^x(x^2-1-2x+2)+C\\&=e^x(x^2-2x+1)+C\\&=e^x(x-1)^2+C\end{align*}

  • Løsningsforslag:\begin{align*}\int (x^2+2x+1)e^x \ dx&=e^x(x^2+2x+1)-\int e^x(2x+2) \ dx\\\text{Delvis integral : }&u=x^2+2x+1 \ , \ u'=2x+2 \\&v=e^x \ , \ v'=e^x \\ -----&-----\\\text{Mellomregning : }\\\int e^x(2x+2) \ dx&=e^x(2x+2)-\int 2e^x \ dx=e^x(2x+2)-2e^x+c\\\text{Delvis integral :}&u=2x+2 \ , \ u'=2\\&v=e^x \ . \ v'=e^x\\-----&-----\\\int (x^2+2x+1)e^x \ dx&=e^x(x^2+2x+1)-(e^x(2x+2)-2e^x)+C\\&=e^x(x^2+2x+1-2x-2+2)+C\\&=e^x(x^2+1)+C\\\end{align*}

Oppgave 3

  • Løsningsforslag:\begin{align*}\int e^x \cdot x^2 \ dx&=x^2e^x-\int 2\cdot x\cdot e^x \ dx\\\text{Delvis integral : }&u=x^2 \ , \ u'=2x \\&v=e^x \ , \ v'=e^x \\-----&-----\\\text{Mellomregning : }\int 2xe^x \ dx&=2xe^x-\int 2e^x \ dx\\&=2e^x(x-1)+C\\-----&-----\\&=x^2e^x-2e^x(x-1)+C\\&=e^x(x^2-2x+2)+C\end{align*}

  • Løsningsforslag:\begin{align*}\int 8x^2\cdot e^{2x} \ dx&=8x^2\cdot \frac{1}{2}e^{2x}-\int 16x\cdot \frac{1}{2}e^{2x} \ dx\\&=4x^2 e^{2x}-8\int x\cdot e^{2x} \ dx\\\text{Delvis integral : }&u=8x^2 \ , \ u'=16 x \\&v=\frac{1}{2}e^{2x} \ , \ v'= e^{2x} \\-----&-----\\ \text{Mellomregning : }\int x\cdot e^{2x} \ dx&=\frac{1}{2}xe^{2x}-\frac{1}{2}\int e^{2x} \ dx=\frac{1}{2}e^{2x}(x-\frac{1}{2})+C\\\text{Delvis integral : }&u=x \ , \ u'=1 \\&v=\frac{1}{2}e^{2x} \ , \ v'=e^{2x}\\-----&-----\\\int 8x^2\cdot e^{2x} \ dx&=4x^2e^{2x}-8(\frac{1}{2}e^{2x}(x-\frac{1}{2}))+C\\&=4x^2e^{2x}-4xe^{2x}+2+C\\&=e^{2x}(4x^2-4x+2)+C\\&=2e^{2x}(2x^2-2x+1)+C\end{align*}