\[ \frac{d}{dx}\left(\frac{x^{\alpha+1}}{\alpha+1}+C\right)=x^\alpha\,. \]

\[ \frac{d}{dx}\,\big(\ln x+C\big)=\frac{1}{x}\,,\quad x\gt 0. \]

\[ \frac{d}{dx}\,\big(e^x+C\big)=e^x\,. \]

\[ \frac{d}{dx}\left(\frac{a^x}{\ln a}+C\right)=\frac{\ln a\cdot a^x}{\ln a}=a^x\,. \]

\[ \frac{d}{dx}\,\big(-\cos x+C)=\sin x. \]

\[ \frac{d}{dx}\,\big(\sin x+C)=\cos x. \]

\[ \frac{d}{dx}\,\big(\cosh x+C) =\sinh x. \]

\[ \frac{d}{dx}\,\big(\sinh x+C) =\cosh x. \] Damit sind alle Integrale bestimmt.\( \qquad\Box \)

\[ \int\limits_{-1}^3x\,dx =\frac{x^2}{2}\,\Big|_{x=-1}^{x=3} =\frac{9}{2}-\frac{1}{2} =\frac{8}{2} =4. \]

\[ \int\limits_0^2(x^3+2x)\,dx =\int\limits_0^2x^3\,dx+2\int\limits_0^2x\,dx =\frac{x^4}{4}\,\Big|_{x=0}^{x=2}+2\,\frac{x^2}{2}\,\Big|_{x=0}^{x=2} =\frac{16}{4}+4 =8. \]

\[ \int\limits_0^1x\sqrt{x}\,dx =\int\limits_0^1x^\frac{3}{2}\,dx =\frac{2x^\frac{5}{2}}{5}\,\Big|_{x=0}^{x=1} =\frac{2}{5}\,. \]

\[ \int\limits_1^2\frac{dx}{2x} =\frac{1}{2}\,\ln x\,\Big|_{x=1}^{x=2} =\frac{1}{2}\,\ln 2-\frac{1}{2}\,\ln 1 =\frac{1}{2}\,\ln 2. \]

\[ \int\limits_0^\pi\cos x\,dx =\sin x\,\Big|_{x=0}^{x=\pi} =\sin\pi-\sin 0 =0. \]

\[ \int\limits_0^3\sinh x\,dx =\cosh x\,\Big|_{x=0}^{x=3} =\cosh 3-\cosh 0 =\cosh 3-1. \]

\[ \int\limits_0^2\frac{3e^x}{2}\,dx =\frac{3}{2}\,\int\limits_0^2e^x\,dx =\frac{3}{2}\,e^x\,\Big|_{x=0}^{x=2} =\frac{3}{2}\,(e^2-e^0) =\frac{3}{2}\,(e^2-1). \]

\[ \int\limits_1^22^x\,dx =\frac{2^x}{\ln 2}\,\Big|_{x=1}^{x=2} =\frac{2^2}{\ln 2}-\frac{2}{\ln 2} =\frac{2}{\ln 2}\,. \] Damit sind alle Integrale ermittelt.\( \qquad\Box \)

\[ \int xe^{-x^2}\,dx =-\,\frac{1}{2}\,\int\frac{d}{dx}\,e^{-x^2}\,dx =-\,\frac{1}{2}\,e^{-x^2}+C\,. \]

\[ \sin^2x=\frac{1}{2}-\frac{1}{2}\,\cos 2x. \]

\[ \int\sin^2x\,dx =\int\frac{dx}{2}-\frac{1}{2}\,\int\cos 2x\,dx =\frac{x}{2}-\frac{1}{4}\,\sin 2x+C. \]

\[ \begin{array}{lll} \displaystyle \int\sin^nx\,dx & = & \displaystyle \int\sin^{n-1}x\cdot\sin x\,dx \\ & = & \displaystyle -\,\sin^{n-1}x\cdot\cos x+(n-1)\int\sin^{n-2}x\cdot\cos^2x\,dx \\ & = & \displaystyle -\,\sin^{n-1}x\cdot\cos x+(n-1)\int\sin^{n-2}x\cdot(1-\sin^2x)\,dx \\ & = & \displaystyle -\,\sin^{n-1}x\cdot\cos x+(n-1)\int\sin^{n-2}x\,dx-(n-1)\int\sin^nx\,dx \end{array} \]

\[ \int\sin^n\,dx=-\,\frac{1}{n}\,\sin^{n-1} x\cdot\cos x+\frac{n-1}{n}\,\int\sin^{n-2}x\,dx. \]

\[ \begin{array}{lll} \displaystyle \int\cos^nx\,dx & = & \displaystyle \int\cos^{n-1}x\cdot\cos x\,dx \\ & = & \displaystyle \cos^{n-1}x\cdot\sin x+(n-1)\int\cos^{n-2}x\cdot\sin^2x\,dx \\ & = & \displaystyle \cos^{n-1}x\cdot\sin x+(n-1)\int\cos^{n-2}x\cdot(1-\cos^2x)\,dx \\ & = & \displaystyle \cos^{n-1}x\cdot\sin x+(n-1)\int\cos^{n-2}x\,dx-(n-1)\int\cos^nx\,dx \end{array} \]

\[ \int\cos^nx\,dx =\frac{1}{n}\,\cos^{n-1}x\cdot\sin x+\frac{n-1}{n}\,\int\cos^{n-2}x\,dx. \] Das war zu zeigen.\( \qquad\Box \)

\[ \begin{array}{lll} \displaystyle \int\limits_0^1(1-3x)^3\,dx & = & \displaystyle -\,\frac{1}{3}\,\int\limits_1^{-2}t^3\,dt \,=\,\frac{1}{3}\,\int\limits_{-2}^1t^3\,dt \\ & = & \displaystyle \frac{1}{3}\cdot\frac{1}{4}\cdot t^4\,\Big|_{t=-2}^{t=1} \,=\,\frac{1}{12}\,(1-16) \,=\,-\,\frac{15}{12} \,=\,-\,\frac{5}{4}\,. \end{array} \]

\[ \int\limits_0^13x^2e^{x^3}\,dx =\int\limits_0^1e^t\,dt =e^t\,\Big|_{t=0}^{t=1} =e-1. \]

\[ \int\limits_0^1e^x\sin e^x\,dx =\int\limits_1^e\sin t\,dt \\ =-\cos t\,\Big|_{t=1}^{t=e} =\cos 1-\cos e. \]

\[ \begin{array}{lll} \displaystyle \int\limits_0^\frac{\pi}{2}\cos(5x+1)\,dx\negthickspace & = & \negthickspace\displaystyle \frac{1}{5}\,\int\limits_1^{\frac{5\pi}{2}+1}\cos t\,dt \,=\,\frac{1}{5}\,\sin t\,\Big|_{t=1}^{t=\frac{5\pi}{2}+1} \\ & = & \negthickspace\displaystyle \frac{1}{5}\,\sin\left(\frac{5\pi}{2}+1\right)-\frac{1}{5}\,\cos 1 \,=\,\frac{1}{5}\,\cos 1-\frac{1}{5}\,\sin 1. \end{array} \]

\[ \int\limits_0^1e^{e^x}e^x\,dx =\int\limits_1^ee^t\,dt =e^t\,\Big|_{t=1}^{t=e} =e^e-e. \]

\[ \int\limits_0^1\frac{x}{\sqrt{2x^2+3}}\,dx =\frac{1}{4}\,\int\limits_3^5\frac{dt}{\sqrt{t}} =\frac{1}{4}\cdot 2\cdot\sqrt{t}\,\Big|_{t=3}^{t=5} =\frac{1}{2}\,(\sqrt{5}-\sqrt{3}). \] Damit sind alle Integrale bestimmt.\( \qquad\Box \)

\[ \begin{array}{lll} \displaystyle \int\limits_0^1xe^x\,dx & = & \displaystyle xe^x\,\Big|_{x=0}^{x=1}-\int\limits_0^1e^x\,dx \\ & = & \displaystyle e-e^x\,\Big|_{x=0}^{x=1} \,=\,e-e+1 \,=\,1. \end{array} \]

\[ \int\limits_0^1x^2e^x\,dx =x^2e^x\,\Big|_{x=0}^{x=1}-2\int\limits_0^1xe^x\,dx =e-2. \]

\[ \int\limits_1^2\ln x\,dx =\int\limits_1^21\cdot\ln x\,dx =x\ln x\,\Big|_1^2-\int\limits_1^2x\cdot\frac{1}{x}\,dx =2\ln 2-\int\limits_1^2dx =2\ln 2-1. \]

\[ \begin{array}{lll} \displaystyle \int\limits_1^2x^2\ln x\,dx\negthickspace & = & \negthickspace\displaystyle \frac{x^3}{3}\,\ln x\,\Big|_1^2-\int\limits_1^2\frac{x^3}{3}\cdot\frac{1}{x}\,dx \,=\,\frac{8}{3}\,\ln 2-\frac{1}{3}\,\int\limits_1^2x^2\,dx \\ & = & \negthickspace\displaystyle \frac{8}{3}\,\ln 2-\frac{x^3}{9}\,\Big|_1^2 \,=\,\frac{8}{3}\,\ln 2-\frac{7}{9}\,. \end{array} \]

\[ \int\limits_0^\pi x\sin x\,dx =-x\cos x\,\Big|_0^\pi+\int\limits_0^\pi\cos x\,dx =-\pi\cos\pi+\sin x\,\Big|_0^\pi \]

\[ \begin{array}{lll} \displaystyle \int\limits_0^\pi e^x\sin x\,dx\negthickspace & = & \negthickspace\displaystyle e^x\sin x\,\Big|_0^\pi-\int\limits_0^\pi e^x\cos x\,dx \,=\,-\int\limits_0^\pi e^x\cos x\,dx \\ & = & \negthickspace\displaystyle -e^x\cos x\,\Big|_0^\pi-\int\limits_0^\pi e^x\sin x\,dx \\ & = & \negthickspace\displaystyle -e^\pi\cos\pi+e^0\cos 0-\int\limits_0^\pi e^x\sin x\,dx \\ & = & \negthickspace\displaystyle e^\pi+1-\int\limits_0^\pi e^x\sin x\,dx \end{array} \]

\[ \int\limits_0^\pi e^x\sin x\,dx=\frac{1+e^\pi}{2}\,. \] Damit sind alle Integrale bestimmt.\( \qquad\Box \)