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               "title": "Primära Funktioner"
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     "Gräsvärde (1).md",
     "Komplexa tal.md",
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+- **Insättning**
+	- **Theorem**: *$F$ är en primitiv funktion till funktionen $f$. Bestämd integralen ges av*$$\int_a^bf(y)dy=F(b)-F(a)$$![[Int1.png]]
+	- 
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diff --git a/Primära Funktioner.md b/Primära Funktioner.md
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 - Regler Example $$\begin{align}\text{Låt }g(x)=y\Rightarrow dy=g'(x)dx\\\int f(g(x))g'(x)dx=\int f(y)dy\\F(x)+C=F(g(x))+C\end{align}$$
 	- **Ex** $$\begin{align}\int\frac1{x^{1/2}+x^{3/2}}dx=I\\\text{Låt }y=\sqrt{x}\Rightarrow dy=\frac1{2\sqrt{x}}dx\\I=\int\frac1{\sqrt{x}\left(1+x\right)}dx=\int\frac2{1+x}\times\frac{dx}{2\sqrt{x}}\\=\int\frac2{1+y^2}dy=2\int\frac1{1+y^2}dy\;\;\left(\frac{d}{dx}\left(\arctan x\right)=\frac1{1+x^2}\right)\\=2\arctan y+C\\=2\arctan\sqrt{x}+C,\text{ där }X\in\mathbb{R}\\\text{prof: }\left(2\arctan\sqrt{x}\right)'\\=\cancel2\times\frac1{1+\left(\sqrt{x}\right)^2}\times\frac1{\cancel2\sqrt{x}}\\=\frac1{x^{1/2}+x^{3/2}}\checkmark\end{align}$$$$\begin{align}\int\cos\left(2x+\pi\right)dx\\=\frac12\sin\left(2x+\pi\right)+C\end{align}$$$$\begin{align}I=\int\frac{\sin x}{\cos x}dx=-\int\frac{\left(\cos x\right)'}{\cos x}dx=-\ln\left|\cos x\right|+C\end{align}$$$$\begin{align}\left(F(x)g(x)\right)'=F'(x)g(x)+F(x)g'(x)\\=f(x)g(x)+F(x)+g'(x)\\F(x)g(x)=\int f(x)g(x)dx+F(x)g'(x)dx\end{align}$$$$\begin{align}\int\left(x^2-4x+5\right)\sin2xdx\\\stackrel{\text{PI}}{=}\left(\int\sin2xdx\right)\left(x^2-4x+5\right)-\int{\left(\int\sin2xdx\right)\left(x^2-4x+5\right)'dx}\\=-\frac12\left(\cos2x\right)\left(x^2-4x+5\right)+\frac12\int\left(\cos2x\right)\left(3x+4\right)dx\\\stackrel{\text{PI}}{=}-\frac12\left(x^2-4x+5\right)\cos2x+\frac12\left(\sin2x\right)\left(x-2\right)-\int\frac12\sin2xdx\\=-\frac12\left(x^2-4x+5\right)\cos2x+\frac12\left(x-2\right)+\frac14\cos2x+C\\=-\frac14\left[\left(2x^2-8x+10-1\right)\cos2x-2(x-2)\sin2x\right]+C\\=\frac{x-2}2\sin2x-\frac{2x^2-8x+9}4\cos2x+C\end{align}$$
 - **Ex kontrol** $$\begin{align}\int\left(\sin x^2\right)\left(2x\right)dx\\=-\cos x^2+C\\\text{prof: }\left(-\cos x^2+C\right)'\\=-\left(-\sin x^2\right)\left(x^2\right)'=\left(\sin x^2\right)\left(x^2\right)\end{align}$$
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+- Rationella Funktioner $$\begin{align}f(x)=\frac{P(x)}{Q(x)}\text{ där }P,Q\text{ är polynomer}\\1.\;\text{ Om }grad(P)\geq grad(Q)\text{ polynomdivition }\\f(x)=\frac{P(x)}{Q(x)}=k(x)+\frac{R(x)}{Q(x)}\\\text{där }K,R\text{ är polynom, }grad(R)<grad(Q)\\2.\;\text{ Faktorisera }Q(x)\\Q(x)=c(x-a_1)(x-a_2)\dots(x^2+b_1x+d)\dots\\3.\;\text{ Partialbråkuppdelning}\\\text{Antag att}\\\frac{R(x)}{Q(x)}=\frac{A_1}{x-a_1}+\frac{A_2}{x-a_2}+\dots\dots\\\text{Bestäm konstanten i HL genom att jämföra med VL}\\4.\;\text{ Integrera}\end{align}$$
+	- **Ex**: $$\begin{align}\int\frac{5x+4}{x^2+3x+2}dx=I\\\text{Lösm: Eftersom }grad(5x+4)<grad(x^2+3x+2)\text{ polinomdivision behövs inte}\\\text{Foktoresera nämnaren: }x^2+3x+2=(x+2)(x+1)\\\text{PBU: Antag att }\exists\text{ konstanten }A,B\in\mathbb{R}:\\\frac{5x+4}{x^2+3x+2}=\frac{A}{x+2}+\frac{B}{x+1}=\frac{A(x+1)+B(x+2)}{(x+2)(x+1)}\\\Rightarrow\;5x+4=A(x+1)+B(x+2)\\\text{Metod 1: Prova olika värde av }x\\x=-1:\;\;5\times(-1)+4=A\times{O}+B(-1+2)\Leftrightarrow B=-1\\x=-2:\;\;5\times(-2)+4=A\times{O}+B\times{O}\Leftrightarrow A=6\\\text{Metod 2: Jämför koefficenten}\\5x+4=A(x+1)+B(x+2)=(A+B)x+(A+2B)\\\text{Jämför koefficienter till }x^n\\\left.\begin{aligned}x^1\;\;:\;\;\;\;5=A+B\\x^0\;\;:\;\;4=A+2B\end{aligned}\right\}\Leftrightarrow\begin{aligned}A=6\\B=-1\end{aligned}\\\frac{5x+4}{x^2+3x+2}=\frac{6}{x+2}-\frac1{x+1}\\\int\frac{5x+4}{x^2+3x+2}dx=\int\left(\frac{6}{x+2}-\frac{1}{x+1}\right)dx\\=6\int\frac1{x+2}dx=\int\frac1{x+1}\\=6\ln\mid{x+2}\mid-\ln\mid{x+1}\mid+C\end{align}$$
+	- **Integral av $\frac{Ax+B}{x^2+ax+b}$** $$\frac{Ax+B}$$
+
+| Faktor i $Q(x)$       | PBU                                                             |
+| --------------------- | --------------------------------------------------------------- |
+| $x-a$                 | $\frac{A}{x-a}$                                                 |
+| $(x-a)^n$             | $\frac{A_1}{x-a}+\frac{A_2}{(x-a)^2}+\dots+\frac{A_n}{(x-a)^n}$ |
+| $(x^2+ax+b),\;a^2<4b$ | $\frac{Ax+B}{x^2+ax+b}$                                         |
+|                       |                                                                 |