"gemischte Integrale" - eine Möglichkeit zu lernen, wann welche Integralregeln Anwendung finden. Also quasi eine Möglichkeit die Aufgaben von "Partielle Integration", "Integration durch Substitution" und "Integration mittels Partialbruchzerlegung" zufällig zu machen. 

Logarithmische Integration

Einfache Integrale lassen sich ohne komplizierte "Tricks" integrieren, indem man auf allgemeine Integrationsregeln und bekannte Grundintegrale zurückgreift.

Einfache Integrale

Die partielle Integration wird verwendet, um eine Funktion zu integrieren, die aus zwei (oder mehreren) Faktoren besteht. Ziel ist es die ursprüngliche Funktion so umzuformen, dass eine neue Funktion entsteht, die einfacher zu Integrieren ist.

Partielle Integration

Bei dieser Integrationsmethode lernst du, Terme in einem Integral durch andere zu ersetzen (Substitution) und so eine schwierige Aufgabe lösbar zu machen.

Integrieren durch Substitution

Integration durch Substitution

Die Partialbruchzerlegung kommt insbesondere bei der Integration von rationalen Funktionen zur Anwendung. Man "zerlegt" das vorlegende Integral in einfache Teilbrüche und kann anschließend mit bekannten Integrationsmitteln weiter machen.

Integration mittels Partialbruchzerlegung

Integration durch Partialbruchzerlegung

Partialbruchzerlegung

Funktionsfolgen

Unterschied In/Homogen

Basis, Basiswechsel

Abbildungsmatrizen

Funktionen bestimmen

Geraden und Ebenen

Gruppen

lineare abbildung 

quadratische ergänzung

Baustatik 1

Anfangswertprobleme :)

Differentialgleichungen

basis 

QR-Algorithmus

QR-Zerlegung

Cholesky-Zerlegung

Bild & Urbild

Zweigstromanalyse

Transistor als Thermometer

Investment

Teilfolgen

Epsilon Delta Kriterium

prinzip der virtuellen verrückung 

Linear Kombination

Folgen, Reihen und Potenzreihen

Flächenpressung

Arbeit 

Cholesky Zerlegung

Polynomfunktionen beschreiben

Zu Maß-und Integrationstheorie: Majorisierte Konvergenz

Teilchenphysik und Festkörperphysik

Orthogonaltrajektorien

test

Taylorreihen

Regel von de L'Hopital

Mathe

KRAFT

Grenzwert Funktion 

Folgenkriterium

ϵ-δ Kriterium 

Beweis Stetigkeit 

Wahrheitstabelle

2x2 Matrix

Matrix invertieren

Basiswechselsatz

Lineare Abbildunge 

Vorderrad fahradgabel lagerkräfte berechnen 

Spannungsregelung

Jordan Normal Form

Test

Induktion 

Jordan-Normalform

Ansatz der Störfunktion zur Bestimmung der Partikulären Lösung

Kurvenintegral

dlg mithilfe inhomogenen linearen system lösen

Inverse einer Matrix

Gleichmäßige Stetigkeit

addition

Regel von L´Hospital

Lineare Abbildungen

dimension eines Untervektorraumes bestimmen

Stationäre Stoffübertragung

Polplan, Prinzip der virtuellen Verrückungen

komplexe Ungleichungen 

Bidualräume

Dualräume

Taylorpolynome

Kern und Basis

Diagonalisieren einer Matrix

Rekursive Folgen

Exponentialreihe

Darstellungsmatrizen

restglied

Basiswechsel

Gamma Funktion

Newton Verfahren

Randwertaufgabe

Numerische Integration: Trapez- und Simpsonregel

Funktionenfolgen

dualräume

Dualraum

Komplexe Nullstellen

Chemie

Natascha Kirchmann

Differentialquotient

Folgedefinition des Funktionslimes

Grenzwerte und Stetigkeit 

Komplexe Zahlen

Kern und Bild 

Direkte Summe 

Koordinaten 

Exponentielle Verzinsung

Lineare Verzinsung

\[\int\frac{x+\frac{1}{2}}{\sqrt{x^2-x+\frac{5}{4}}}\text{d}x\space =\sqrt{x^2-x+\frac{5}{4}}\space+\arctan(x-\frac{1}{2})+c,\quad c\in\mathbb{R}\]

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Bringe den Term unter der Wurzel zuerst in Scheitelpunktsform:

\[x^2-x+\frac{5}{4} = x^2-x+\frac{1}{4}-\frac{1}{4}+\frac{5}{4}$
                     $=(x-\frac{1}{2})^2+1\]

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Substituiere: $u:= x-\frac{1}{2}\to \text{d}u = -\frac{1}{2}\space\text{d}x$

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\[\to\int\frac{u+1}{\sqrt{u^2+1}}\]

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\[\int\frac{u}{\sqrt{u^2+1}}\text{d}u\space +\int{\frac{1}{\sqrt{u^2+1}}}\text{d}u\]

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Substituiere: $t:=u^2+1\to \text{d}t = 2u\space\text{d}u$

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\[\to\int\frac{1}{2}\frac{1}{\sqrt{t}}\text{d}t\space+\int\frac{1}{\sqrt{u^2+1}}\text{d}u\]

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Berechne das erste Integral mittels der Formel:
\[\int x^n\text{d}x=\frac{1}{n+1}x^{n+1}+c\]

Berechne das erste Integral mittels der Formel:
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Das zweite Integral ist ein Standardintegral:
$\int\frac{1}{x^2+1}\text{d}x=\arctan(x)+c$

Das zweite Integral ist ein Standardintegral:
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\[\int\frac{1}{2}\frac{1}{\sqrt{t}}\text{d}t\space+\int\frac{1}{\sqrt{u^2+1}}\text{d}u\space=\sqrt{t}+\arctan(u)+c\]

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\[\to\sqrt{u^2+1}+\arctan(x-\frac{1}{2})+c\]
\[\to\sqrt{x^2-x+\frac{5}{4}}\space+\arctan(x-\frac{1}{2})+c,\quad c\in\mathbb{R}\]

Man berechne: 
\[\int\frac{x+\frac{1}{2}}{\sqrt{x^2-x+\frac{5}{4}}}\text{d}x\]

Man berechne: 
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Dies ist eine interaktive Aufgabe zu: Integrieren durch Substitution mit praktischen Tipps zum Lösen und einer Zusammenfassung der nötigen Theorie

Integrieren - Integrieren durch Substitution | Aufgabe mit Lösung

<h2 class="content_sub_heading">Integrationsregeln im Überblick</h2>
Seien $\alpha \text {,} \ c \in \mathbb{R}$ , konst. und $n \in \mathbb{R} $, $n \neq -1$
Faktorregel
$$\quad \int \alpha f(x) d x=\alpha \cdot \int f(x) d x$$
Summenregel
$$\quad \int(u(x)+v(x)) d x=\int u(x) d x+\int v(x) d x$$
Konstantenregel
$$\quad \int \alpha \cdot d x= \alpha \cdot x+c$$
Potenzregel
$$\quad \int x^{n} d x=\frac{1}{n+1} x^{n+1}+c$$
Partielle Integration
$$\quad \int f(x) g^{\prime}(x) d x=f(x) g(x)-\int f^{\prime}(x) g(x) d x$$
Integration durch Substitution
$$\quad \int f(x) \mathrm{d} x=\int f(g(z)) \cdot g^{\prime}(z) \mathrm{d} z$$

<h2 class="content_sub_heading">Integration durch Substitution</h2>
<ol>
<li>Wähle einen Term aus, den du durch $z$ ersetzen willst: $z=z(x)=g(x)$ </li>
<li>Bestimme $d x$ durch Ableiten von $z(x)=g(x)$ und anschließendem umformen: \begin{align*}\quad \frac{d z}{d x}=g^{\prime}(x) \Rightarrow d x=\frac{d z}{g^{\prime}(x)}\end{align*} </li>
<li>Bestimme neue Integralgrenzen, durch einsetzen von $a, b$ in das in Schritt 1. gewählte $z(x)$: $\quad \Rightarrow g(a)=\ldots$ und $g(b)=\ldots$ Falls es sich um ein unbestimmtes lntegral (lntegral ohne Grenzen $a \ \text{und} \ b$) handelt, diesen Schritt weglassen!   </li>
<li>Ersetze nun jeden Term $g(x)$ durch $z$, jedes $d x$ durch $\frac{d z}{g^{\prime}(z)}$ und (falls vorhanden) die Integrationsgrenzen $a, b$ durch $g(a) \ \text{und} \ g(b)$. Das neue Integral sollte nun kein $x$ mehr enthalten: \begin{align*}\quad \int_{a}^{b} \ldots d x \quad \Rightarrow \quad \int_{g(a)}^{g(b)} \ldots d z\end{align*} </li>
<li>Integriere den neuen Ausdruck mithilfe der Integrationsregeln. </li>
<li>Falls ein unbestimmtes Integral (Integral ohne Grenzen) vorlag, so musst du noch resubstituieren. Ersetze hierfür jedes $z$ wieder durch $g(x)$.</li>
</ol>

<h2 class="content_sub_heading">Liste wichtiger Grundintegrale</h2>
Potenzen:
$\quad \int x^{n} d x =\frac{x^{n+1}}{n+1} \quad \text { mit } n \neq-1$
$\quad \int \frac{1}{x} dx =\ln |x|$
 
Gebrochenrationale Funktionen:
$\quad \int \frac{d x}{1+x^{2}}=\arctan x$
$\quad \int \frac{d x}{1-x^{2}}=\operatorname{artanh} x=\frac{1}{2} \ln \frac{1+x}{1-x} \quad \text{ mit } |x| \leq 1$
 
Exponentialfunktionen:
$\quad \int e^{x} d x=e^{x}$
$\quad \int a^{x} d x=\frac{a^{x}}{\ln a}$
 
Irrationale Funktionen:
$\quad \int \frac{d x}{\sqrt{1-x^{2}}} \ \ \ \ =\arcsin x$
$-\int \frac{d x}{\sqrt{1+x^{2}}} =\arccos x$
$\quad \int \frac{d x}{\sqrt{1+x^{2}}} \ \ \ \ =\operatorname{arsinh} x=\ln (x+\sqrt{x^{2}+1})$
$\quad \int \frac{d x}{\sqrt{x^{2}-1}} \ \ \ \ =\operatorname{arcosh} x=\ln (x+\sqrt{x^{2}-1})$
 
Trigonometrische Funktionen:
$\quad \int \sin x \ d x \ =-\cos x$
$\quad \int \cos x \ d x=\sin x$
$\quad \int \tan x \ d x=-\ln |\cos x|$
$\quad \int \cot x \ d x=\ln |\sin x|$
$\quad \int \frac{d x}{\cos ^{2} x} \ \ \ \ \ =\tan x$
$\quad \int \frac{d x}{\sin ^{2} x} \ \ \ \ \ =-\cot x$
 
Hyperbelfunktionen:
$\quad \int \sinh x \ d x=\cosh x$
$\quad \int \cosh x \ d x=\sinh x$
$\quad \int \tanh x \ d x=\ln \cosh x$
$\quad \int \operatorname{coth} x \ d x=\ln |\sinh x|$
$\quad \int \frac{d x}{\cosh ^{2} x} \ \ \ \ \ =\tanh x$
$\quad \int \frac{d x}{\sinh ^{2} x} \ \ \ \ \ \ =-\operatorname{coth} x$

Aufgabenstellung:

Lösungsweg:

Lösung: