\(\int\limits_{-5}^{-1} \frac 1{4x-3} \ \mathrm{d}x= \left. \frac 14 \ln(4x-3) \right|_{-5}^{-1}=\)

\(\frac 14 [\ln|-7|-\ln|-23|]= \frac 14 \ln\frac 7{23}= -0,297\)

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

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

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

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

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

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

\(\int\limits_{4}^{8}\frac{1}{\sqrt{2x-7}} \ \mathrm{d}x= \int\limits_{4}^{8}( 2x-7)^{-\frac 12} \ \mathrm{d}x= \)

\(\left. \frac 12 \cdot \frac{(2x-7)^{\frac 12}}{\frac 12} \right|_4^8= 9^{\frac 12}-1^{\frac 12}=2\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\int\limits_{4}^{8}\frac{1}{(x-3)^2} \ \mathrm{d}x= \int\limits_{4}^{8}(x-3)^{-2} \ \mathrm{d}x= \)

\(\left. \frac{(x-3)^{-1}}{-1} \right|_4^8= \left. -\frac 1{x-3} \right|_4^8= -\frac 15 + 1 = \frac 45\)

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

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

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

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

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