\(\int (3x+5)^4 \ \mathrm{d}x\)

Setze  \(u= 3x+5\)

\(\frac{\mathrm{d}u}{\mathrm{d}x}= 3 \\ \mathrm{d}x=\frac 1 3 \mathrm{d}u\) 

\(\int (3x+5)^4 \ \mathrm{d}x = \int u^4 \frac 1 3 \cdot \mathrm{d}u = \frac 1 3 \cdot \frac{u^5}{5}=\\ \frac 1 {15} (3x+5)^5 + C\)

\(\int (-5x+8)^7 \ \mathrm{d}x \)

\(\text{setze } -5x+8=u\)

\(-5= \frac {\mathrm{d}u}{\mathrm{d}x} \)

\(\mathrm{d}x=-\frac 1 5 \mathrm{d}u\)

\(\int (-5x+8)^7 \ \mathrm{d}x =\int u^7 \left(-\frac 1 5 \right) \mathrm{d}u= -\frac 1 5 \int u^7 \mathrm{d}u = \)

\(-\frac 1 5 \cdot \frac {u^8}{8} = -\frac 1 {40} (-5x+8)^8 +C\)

\(\int \frac 1 {3x+4} \ \mathrm{d} x\)

\(\text{setze } 3x+4=u\)

\(3=\frac {\mathrm{d}u} {\mathrm{d}x}\)

\(\mathrm{d}x=\frac 1 3 \mathrm{d}u\)

\(\int \frac 1 {3x+4} \ \mathrm{d} x = \int \frac 1 u \cdot \frac 1 3 \cdot \mathrm{d}u=\)

\(\frac 1 3 \ln u = \frac 1 3 \ln(3x+4)+C\)

\(\int \frac 1 {(3x+4)^2} \ \mathrm{d}x\)

\(\text{setze } 3x+4=u\)

\(3=\frac{\mathrm{d} u}{\mathrm{d}x}\)

\(\mathrm{d}x=\frac 1 3 \mathrm{d}u\)

\(\int \frac 1 {(3x+4)^2} \ \mathrm{d}x = \int \frac 1 {u^2} \cdot \frac 1 3 \cdot \mathrm{d}u = \frac 1 3 \cdot \int u^{-2} \ \mathrm{d}u=\)

\(\frac 1 3 \cdot \frac {u^{-1}}{-1}=-\frac 1 3 \cdot \frac 1 u = -\frac 1 {3(3x+4)}+C\)

\(\int A\cdot \sin(\omega t+\varphi) \ \mathrm{d}t\)

\(\text{setze } \ \omega t + \varphi = u\)

\(\omega=\frac {\mathrm{d}u}{\mathrm{d}t}\)

\(\mathrm{d}t=\frac 1 \omega \mathrm{d}u\)

\(\int A\cdot \sin(\omega t+\varphi_0) \ \mathrm{d}t = \frac A \omega \cdot \int \sin u \cdot \mathrm{d}u=\)

\(\frac A \omega (-\cos u) = -\frac A \omega \cos(\omega t + \varphi)+C\)

\(\text{setze } u=2x\)

\(\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x} 2x\)

\(\mathrm{d}x= \frac 12 \mathrm{d}u\)

\(\int e^{2x}\ \mathrm{d}x= \int e^u \cdot \frac 12\cdot \mathrm{d}u=\)

\(\frac 12 e^u = \frac{e^{2x}}2 +C\)

\(\text{setze } u=3x-5\)

\(\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(3x-5)\)

\(\mathrm{d}x=\frac 13 \mathrm{d}u\)

\(\int e^{3x-5}\ \mathrm{d}x= \int e^u \cdot \frac 13 \mathrm{d}u =\)

\(\frac 13 \cdot \int e^u \mathrm{d}u = \frac 13 e^u= \frac{e^{3x-5}}{3}+C\)

\(\text{setze } u=-2x+7\)

\(\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(-2x+7)\)

\(\mathrm{d}x=-\frac 12 \mathrm{d}u\)

\(\int e^{-2x+7}\ \mathrm{d}x= \int e^u \cdot \left( -\frac 12\right) \mathrm{d}u =\)

\(\left( -\frac 12\right) \cdot \int e^u \mathrm{d}u = \left( -\frac 12\right) e^u= -\frac{e^{-2x+7}}{2}+C\)

\(\text{setze } u=-\frac t\tau\)

\(\frac{\mathrm{d}u}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}(-\frac t\tau)\)

\(\mathrm{d}t=-\tau \ \mathrm{d}u\)

\(\int U \cdot e^{-\frac t\tau}\ \mathrm{d}t= \int U \cdot e^{u}\cdot (-\tau)\ \mathrm{d}u=\)

\( U \cdot e^{u}\cdot (-\tau)\ = -U\tau e^{-\frac t\tau}+C\)

\(\int x^3 \cdot e^{x^4} \ \mathrm{d}x\)

\(\text{setze } u=x^4\)

\(\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(x^4)\)

\(\mathrm{d}x=\frac {1}{4x^3} \mathrm{d}u\)

\(\int x^3 \cdot e^{x^4} \ \mathrm{d}x= \int x^3 \cdot e^u \cdot \frac{1}{4x^3} \mathrm{d}u =\)

\(\left( \frac 14 \right) \cdot \int e^u \mathrm{d}u = \left( \frac 14 \right) \cdot e^u = \)

\(\left( \frac 14 \right) \cdot e^{x^4} + C \)

\(\int \frac{e^x}{1-e^x}\ \mathrm{d}x\)

\(\text{setze } u=1-e^x\)

\(\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(1-e^x)\)

\(\mathrm{d}x=- \frac {1}{e^x} \mathrm{d}u\)

\(\int \frac{e^x}{1-e^x}\ \mathrm{d}x= \int \frac{e^x}{u}\cdot ( - \frac{1}{e^x}) \mathrm{d}u =\)

\(- \int \left( \frac 1u \right) \cdot \mathrm{d}u = - \ln\| u \| + C =- \ln\| 1-e^x \| + C = \)

\( \ln\| (1-e^x)^{-1} \| + C = \ln\| \frac{1}{(1-e^x)}\| + C \)

\(\int \frac{1}{x \cdot \ln(x) } \ \mathrm{d}x\)

\(\int x^3 \cdot e^{x^4} \ \mathrm{d}x\)

\(\text{setze } u=\ln(x)\)

\(\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(\ln(x))\)

\(\mathrm{d}x=x\cdot \mathrm{d}u\)

\(\int \frac{1}{x \cdot \ln(x) } \ \mathrm{d}x = \int \frac{1}{x \cdot u } \cdot x \ \mathrm{d}u = \)

\(\int \frac{1}{u } \ \mathrm{d}u = \ln(u) + C = \ln(\ln(x)) + C\)

\(\int x \cdot \cos(x^2) \ \mathrm{d}x\)

\(\text{setze } u=x^2\)

\(\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(x^2)\)

\(\mathrm{d}x=\frac {1}{2x} \mathrm{d}u\)

\(\int x \cdot \cos(x^2) \ \mathrm{d}x = \int x \cdot \cos(u) \cdot \frac1{2x} \ \mathrm{d}u=\)

\(\frac12 \cdot \int \cos(u) \ \mathrm{d}u = \frac12 \cdot \sin(u) \ = \)

\(\frac12 \cdot \sin(x^2)\)

\(\int x \cdot \sin(x^2) \ \mathrm{d}x\)

\(\text{setze } u=x^2\)

\(\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(x^2)\)

\(\mathrm{d}x=\frac {1}{2x} \mathrm{d}u\)

\(\int x \cdot \sin(x^2) \ \mathrm{d}x = \int x \cdot \sin(u) \cdot \frac1{2x} \ \mathrm{d}u=\)

\(\frac12 \cdot \int \sin(u) \ \mathrm{d}u = - \frac12 \cdot \cos(u) \ = \)

\(-\frac12 \cdot \cos(x^2)\)

\(\int \frac{1}{\sqrt{x-1}} \ \mathrm{d}x\)

\(\text{setze } u=x-1\)

\(\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(x-1)\)

\(\mathrm{d}x= \mathrm{d}u\)

\(\int \frac{1}{\sqrt{x-1}} \ \mathrm{d}x = \int \frac{1}{\sqrt{u}} \ \mathrm{d}u = \)

\(\int u^{-\frac12}\ \mathrm{d}u = u^{\frac12}\ \cdot \frac1{\frac12} + C = \)

\(u^{\frac12}\ \cdot\ 2 + C= 2 \cdot (x-1) + C= 2 \cdot \sqrt{x-1}+ C\)

\(\int \frac{x}{\sqrt{100-2x^2}} \ \mathrm{d}x\)

\(\text{setze } u=100-2x^2\)

\(\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(100-2x^2)\)

\(\mathrm{d}x= -\frac1{4x}\mathrm{d}u\)

\(\int \frac{x}{\sqrt{100-2x^2}} \ \mathrm{d}x = \int \frac{x}{\sqrt{u}} \ \cdot (-\frac1{4x}) \mathrm{d}u = \)

\(-\frac14 \cdot \int \frac{1}{\sqrt{u}} \ \mathrm{d}u = -\frac14 \cdot \int u^{-\frac12} \ \mathrm{d}u = \)

\(-\frac14 \cdot u^{\frac12} \cdot \frac1{\frac12}} = -\frac14 \cdot 2 u^{\frac12} = \)

\(- \frac12 \cdot (100-2x^2)^{\frac12} = - \frac12 \sqrt {100-2x^2}\)

\(\text{setze } u=-x\)

\(\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(-x)\)

\(\mathrm{d}x=- \mathrm{d}u\)

\(\int e^{-x}\ \mathrm{d}x = - \int e^u \cdot \mathrm{d}u =\)

\(-e^u + C = - e^{-x} + C \)

\(\int \frac{x}{k^2 + x^2}\ \mathrm{d}x\)

\(\text{setze } u=k^2 + x^2\)

\(\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(k^2 + x^2)\)

\(\mathrm{d}x= \frac {1}{2x} \mathrm{d}u\)

\(\int \frac{x}{k^2 + x^2}\ \mathrm{d}x = \int \frac{x}{u}\ \frac {1}{2x} \mathrm{d}u = \)

\( \int \frac{1}{2u}\ \mathrm{d}u = \frac 12 \cdot \ln \| u \| + C = \frac 12 \cdot \ln \| k^2 + x^2 \| + C\)

\(\int \frac{ \ln(x)}{x } \ \mathrm{d}x\)

\(\text{setze } u=\ln(x)\)

\(\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(\ln(x))\)

\(\mathrm{d}x=x\cdot \mathrm{d}u\)

\(\int \frac{ \ln(x)}{x } \ \mathrm{d}x = \int \frac{u }{x } \cdot x \ \mathrm{d}u = \int u \ \mathrm{d}u = \)

\(\frac{u^2}2 + C = \frac{\(ln(x))^2}2 + C \)

\(\int \frac{\sin(x)}{1-\cos(x)} \ \mathrm{d}x\)

\(\text{setze } u=\cos(x)\)

\(\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(\cos(x))\)

\(\mathrm{d}x=-\frac1{\sin(x)}\cdot \mathrm{d}u\)

\(\int \frac{\sin(x)}{1-\cos(x)} \ \mathrm{d}x = \int \frac{\sin(x)}{1-u} \cdot (-\frac1{\sin(x)}) \mathrm{d}u = - \int \frac1{1-u} \mathrm{d}u =\)

\(\text{setze } 1-u=t\)

\(\frac{\mathrm{d}t}{\mathrm{d}u}=\frac{\mathrm{d}}{\mathrm{d}u}(1-u)\)

\(\mathrm{d}u=- \mathrm{d}t\)

\( - \int \frac1{1-u} \mathrm{d}u = -\int \frac1{t} \cdot (-1) \mathrm{d}t = \int \frac1{t} \cdot \mathrm{d}t = \ln \|t \| + C = \)

\( \ln \|1-u \| + C = \ln \|1 - \cos(x) \| + C \)

\(\int \frac1{x^2-1} \ \mathrm{d}x = \int \frac1{(x+1)(x-1)} \ \mathrm{d}x\)

Zerlegen des Terms mit Hilfe der Partialbruchzerlegung

\( \frac1{(x+1)(x-1)} = \frac A{x+1} + \frac B{x-1}\)

\( \frac1{(x+1)(x-1)} = -\frac12 \cdot \frac 1{x+1} + \frac12 \cdot \frac 1{x-1}\)

\( \int \frac1{(x+1)(x-1)} \ \mathrm{d}x = -\int \frac12 \cdot \frac 1{x+1}\mathrm{d}x +\int \frac12 \cdot \frac 1{x-1} \ \mathrm{d}x\)

\(\text{setze } u=x+1 \vspace \text{und} \vspace t = x-1 \)

\(-\frac12 \ln \|u \| + \frac12 \ln \| t \| + C = -\frac12 \ln \|x+1 \| + \frac12 \ln \| x-1 \| + C =\)

\(\frac12 \ln \| \frac{x-1}{x+1} \| + C\)

\( \int \! \sin (nx) \ \mathrm{d}x \)

\(\text{setze } u=nx\)

\(\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(nx)\)

\(\mathrm{d}x=\frac {1}{n} \mathrm{d}u\)

\( \int \! \sin (nx) \ \mathrm{d}x = \frac1n \int \! \sin (u) \ \mathrm{d}u = -\frac1n \cos(u) = -\frac1n \cos(nx)\)

\( \int \! \cos (nx) \ \mathrm{d}x \)

\(\text{setze } u=nx\)

\(\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(nx)\)

\(\mathrm{d}x=\frac {1}{n} \mathrm{d}u\)

\( \int \! \cos (nx) \ \mathrm{d}x = \frac1n \int \! \cos (u) \ \mathrm{d}u = \frac1n \sin(u) = \frac1n \sin(nx)\)

\(\int x \cdot e^{x^2} \mathrm{d}x \)

\(\text{setze } u=x^2\)

\(\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(x^2)\)

\(\mathrm{d}x=\frac {1}{2x} \mathrm{d}u\)

\(\int x \cdot e^{x^2} \mathrm{d}x = \int x \cdot e^{u} \frac1{2x} \mathrm{d}u = \int \frac12 e^u \mathrm{d}u = \frac12 e^u = \frac12 e^{x^2} \)

\( \int \! \tan (nx) \ \mathrm{d}x = \int \! \frac{\sin(nx)}{\cos(nx)} \ \mathrm{d}x \)

\(\text{setze } u=\cos(nx)\)

\(\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(\cos(nx))\)

\(\mathrm{d}x=-\frac {1}{n\sin(x)} \mathrm{d}u\)

\( \int \! \frac{\sin(nx)}{\cos(nx)} \ \mathrm{d}x = - \int \! \frac{\sin(nx)}{ u} \cdot \frac1{n\sin(nx)} \ \mathrm{d}u = \)

\(- \frac1n \int \frac1u \mathrm{d}u = -\frac1n \ln|u| = -\frac1n \ln|\cos(nx)|\)

\( \int \! \tan (x) \ \mathrm{d}x = \int \! \frac{\sin(x)}{\cos(x)} \ \mathrm{d}x \)

\(\text{setze } u=\cos(x)\)

\(\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(\cos(x))\)

\(\mathrm{d}x=-\frac {1}{n\sin(x)} \mathrm{d}u\)

\( \int \! \frac{\sin(x)}{\cos(x)} \ \mathrm{d}x = - \int \! \frac{\sin(x)}{ u} \cdot \frac1{\sin(x)} \ \mathrm{d}u = \)

\(- \int \frac1u \mathrm{d}u = - \ln|u| = - \ln|\cos(x)|\)