Integration durch Substitution

Partielle Integration 

Integralgleichung

kettenregel

Lösbarkeit einer Matrix

Glechgewicht

Schiefe Ebene

Elektrische Anlage mit Abzweigunen

Flächenträgheit

Wechselstrom

Cauchy Verdichtungssatz 

Schaltvorgänge

Singulärwertzerlegung

Spann

Lineares Gleichungssystem

spannung

LR-Zerlegung mit pivotisierung 

Basis bestimmen

trigonalisierbarkeit

Fourier Koeffizienten bestimmen

Reimann-Summen

Ungleichungen 

Statik in der Ebene

Fourier-Reihen

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

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

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

Substituiere: <mjx-container class="MathJax" jax="SVG"><svg alt="t:=u^2+1\to \text{d}t = 2u\space\text{d}u" style="vertical-align: -0.186ex;" xmlns="http://www.w3.org/2000/svg" width="24.796ex" height="2.072ex" role="img" focusable="false" viewBox="0 -833.9 10959.7 915.9" xmlns:xlink="http://www.w3.org/1999/xlink"><defs><path id="MJX-15-TEX-I-1D461" d="M26 385Q19 392 19 395Q19 399 22 411T27 425Q29 430 36 430T87 431H140L159 511Q162 522 166 540T173 566T179 586T187 603T197 615T211 624T229 626Q247 625 254 615T261 596Q261 589 252 549T232 470L222 433Q222 431 272 431H323Q330 424 330 420Q330 398 317 385H210L174 240Q135 80 135 68Q135 26 162 26Q197 26 230 60T283 144Q285 150 288 151T303 153H307Q322 153 322 145Q322 142 319 133Q314 117 301 95T267 48T216 6T155 -11Q125 -11 98 4T59 56Q57 64 57 83V101L92 241Q127 382 128 383Q128 385 77 385H26Z"></path><path id="MJX-15-TEX-N-3A" d="M78 370Q78 394 95 412T138 430Q162 430 180 414T199 371Q199 346 182 328T139 310T96 327T78 370ZM78 60Q78 84 95 102T138 120Q162 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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: Integration durch Substitution mit praktischen Tipps zum Lösen und einer Zusammenfassung der nötigen Theorie

Spezielle Integrationstechniken - Integration durch Substitution | Auf

<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: