\(\int\limits_{0}^{10^6} e^{-x} \mathrm{d}x= \left. \frac{e^{-x}}{-1}\right|_0^{10^6}= e^{-x} \ |_0^{10^6}= -e^{-10^6}-(-e^0)\approx 0+1=1\)

\(\text{setze } u=-\frac t3 \)

\(\frac{\mathrm{d}}{\mathrm{d}t} (-\frac t3) = \frac{\mathrm{d}u}{\mathrm{d}t}\)

\(dt = -3\ \mathrm{d}u\)

\(\left. \int\limits_{-2}^{5} e^{-\frac t3} \mathrm{d}t= -3 \int\limits_{-2}^{5} e^u \ \mathrm{d}u = -3\cdot e^u \right|_{-2}^5=\)

\( \left. -3\cdot e^{-\frac t3}\ \right|_{-2}^5= -3\left( e^{-\frac 53}-e^{\frac 23} \right) =5,28\)

\(\int\limits_{2}^{7} (2x-1)^3 \ \mathrm{d}x= \left. \frac 12 \cdot \frac{(2x-1)^4}{4}\ \right|_2^7 =\)

\(\frac 18 \left[ (2\cdot 7 - 1)^4-(2\cdot2-1)^4\right] = \frac 18 \left( 13^4-3^4\right) =3560\)

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

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

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

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

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

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

\(\int\limits_{0}^{2} \cos\left(\frac{2\pi}{8}x\right) \ \mathrm{d}x= \left. \frac{8}{2\pi}\cdot \sin\left(\frac{2\pi}{8}x\right) \ \right|_0^2=\)

\(\frac 4\pi \left[ \sin\left( \frac{2\pi}{8}2\right) -\sin\left( \frac{2\pi}{8}\cdot0\right) \right]= \frac 4\pi \sin\left( \frac{\pi}{2}\right)=\frac 4\pi\)

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

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

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

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

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

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

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

\(\int\limits_{0}^{\sqrt\pi} x \cdot \cos(x^2) \ \mathrm{d}x = [\frac12 \cdot \sin(x^2)]_{0}^{\sqrt\pi} = \)

\(\frac12 \cdot \sin(\pi) - \frac12 \cdot \sin(0) = 0\)

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

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

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

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

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

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

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

\(\int\limits_{0}^{\frac{\sqrt\pi}2} x \cdot \sin(x^2) \ \mathrm{d}x = [-\frac12 \cdot \cos(x^2)]_{0}^{\frac{\sqrt\pi}2} = \)

\(-\frac12 \cdot \cos((\frac{\sqrt\pi}{2})^2) - (-\frac12 \cdot \cos(0^2)) = \)

\(-\frac12 \cdot \cos(\frac{\pi}{4}) + \frac12 \cdot 1 = \)

\(-\frac12 \cdot \frac{\sqrt 2}{2}+ \frac12 = \)

\(\frac{2-\sqrt2}{4}\)

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

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

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

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

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

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

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

\( \int\limits_{1}^{2} \frac{e^x}{1-e^x}\ \mathrm{d}x = [ \ln\| \frac{1}{(1-e^x)}\| ]_{1}^{2} = \)

\( [ \ln\| \frac{1}{(1-e^2)}\| - \ln\| \frac{1}{(1-e)}\| ]=\)

\( \ln\| \frac{1}{(1-e^2)} : \frac{1}{(1-e)} \| = \ln\| \frac{1}{(1-e^2)} \cdot(1-e) \| = \)

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

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

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

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

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

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

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

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

\(\int \limits_{e}^{e^2} \frac{1}{x \cdot \ln(x) } \ \mathrm{d}x = [\ln(\ln(x)) ]_{e}^{e^2} = \)

\(\ln(\ln(e^2)) - \ln(\ln(e)) = \ln(2) - \ln(1) = \ln(2) \)

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

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

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

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

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

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

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

\(\lim\limits_{a \to 1} \int \limits_{a}^{e} \frac{1}{x \cdot \ln(x) } \ \mathrm{d}x = \lim\limits_{a \to 1} [\ln(\ln(x)) ]_{a}^{e} = \)

\(\lim\limits_{a \to 1} [\ln(\ln(e)) - \ln(\ln(a)) ] = 0 - (-\infty) = \infty\)

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

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

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

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

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

\(\lim\limits_{a \to \infty} \int \limits_0^a e^{-x}\ \mathrm{d}x = \lim\limits_{a \to \infty} [- e^{-x} ]_0^a = \)

\(\lim\limits_{a \to \infty} [-\frac{1}{e^x} ]_0^a = -\frac{1}{e^\infty} - (-\frac{1}{e^0}) = 0 + 1 = 1\)

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

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

\(\mathrm{d}x=- \frac12 \mathrm{d}u\)

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

\(- \frac12 e^u = - \frac12 e^{-2x} \)

\(\lim\limits_{a \to - \infty} \int \limits_a^0 e^{-2x}\ \mathrm{d}x = \lim\limits_{a \to - \infty} [- \frac12 e^{-2x} ]_a^0 = \)

\(\lim\limits_{a \to - \infty} [- \frac12 \cdot \frac{1}{e^{2x}} ]_a^0 = \)

\(- \frac12 \cdot \frac{1}{e^{2\cdot 0 }} - (- \frac12 \cdot \frac{1}{e^{- \infty}}) = \)

\(- \frac12 \cdot 1 + \frac12 \cdot e^{ \infty} = - \frac12 + e^{ \infty} = e^{ \infty} = \inft\)

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

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

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

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

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

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

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

\(\int \limits_1^5 \frac{1}{\sqrt{x-1}} \ \mathrm{d}x = [2 \cdot \sqrt{x-1}]_1^5 =\)

\(2 \cdot \sqrt{4} - 2 \cdot \sqrt{0} = 4\)

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

\(\left. \frac{(x-3)^{-1}}{-1} \right|_4^\infty= \left. -\frac 1{x-3} \right|_4^\infty= \lim\limits_{a \to \infty} -\frac 1{a-3} + 1 = 0+1 = 1\)

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

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

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

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

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

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

\(\int \limits_{1}^{e^2}\frac{ \ln(x)}{x } \ \mathrm{d}x = [\frac{\(ln(x))^2}2 ]_{1}^{e^2}\)

\(\frac{\(ln(e^2))^2}2 - \frac{\(ln(1))^2}2 = \frac42 - 0 = 2\)

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

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

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

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

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

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

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

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

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

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

\(\int\limits_{\frac\pi4}^{\frac\pi2} \frac{\sin(x)}{1-\cos(x)} \ \mathrm{d}x 0 = [\ln \|1 - \cos(x) \| ]_{\frac\pi4}^{\frac\pi2} = \)

\(\ln \|1 - \cos(\frac\pi2) \| - \ln \|1 - \cos(\frac\pi4) \| = \ln \|1 \| - \ln \|1 - \frac{\sqrt2}2 \| = - \ln \|\frac{2 - \sqrt2}2 \|\)

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

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

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

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

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

\(\int \limits_{0}^\pi \! \sin (nx) \ \mathrm{d}x = [ -\frac1n \cos(nx)]_{0}^\pi = - \frac1n \cos(n\pi)+\frac1n \cos(0) + = \frac1n - \frac1n \cos(n\pi) \)

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

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

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

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

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

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

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

\(\int \limits_{0}^{\pi} x \cdot \cos(x^2) \ \mathrm{d}x = [\frac12 \cdot \sin(x^2)]_{0}^{\pi} = [\frac12 \cdot \sin(\pi^2)]- [\frac12 \cdot \sin(0)] = \frac12 \cdot \sin(\pi^2)\)

\(\int\limits_0^1 x \cdot e^{x^2} \mathrm{d}x \)

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

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

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

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

\(\int \limits_0^1 x \cdot e^{x^2} \mathrm{d}x = [\frac12 e^{x^2}]_0^1 = \frac12 e^{1} - \frac12 e^{0} = \frac{e-1}2\)