lineare-substitution-und-integrale

Bsp:Lineares Substituioren $\int \sin (3x-1)dx$ $=\int \sin u\ast x^2\ast \frac{du}{3x^2}$ $\int \frac{1}{3}\sin udu=-\frac{1}{3}\cos u+c$ $=-\frac{1}{3}\cos (x^3)+c$ $u=3x-1$ $\frac{du}{dx}=3$ $dx=\frac{du}{3}$

6.38) a) $\int \frac{2}{2+x}dx$

6.37) b) $\int (3-2x{)}^{4}dx$ c) $\int \sqrt{1-x}dx$ 6.45) a) $\int \frac{x}{(x^2-1{)}^2}dx;u=x^2+1$ c) $\int \frac{\cos x}{{\sin }^{2}x}dx;u=\sin x$ 6.37) d) $\int \frac{1}{\sqrt{1-x}}dx=\int (1-x{)}^{-1/2}dx$ $u=1-x$ → ableiten $\frac{du}{dx}=-1\Rightarrow dex=\frac{du}{-1}=dx=-du$ $\int u^{-1/2}(-du)=-2\ast u^{1/2}+c=-2\ast \sqrt{1-x}+c$ e) $\int (\frac{x}{2}-2{)}^{2}dx$ $u=\frac{x}{2}-2$ $\frac{du}{dx}=\frac{1}{2}$ $du=\frac{1}{2}\ast dx|:\frac{1}{2}|\ast 2$ $dx=2\ast du$ $\int u^2\ast 2\ast du=2\ast \int u^2\ast du=2\ast \frac{u^3}{3}+c$ $\frac{2}{3}\ast (\frac{x}{2}-2{)}^3+c$