Integral von $$$\frac{x^{5} - 1}{x^{3}}$$$
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Ihre Eingabe
Bestimme $$$\int \frac{x^{5} - 1}{x^{3}}\, dx$$$.
Lösung
Expand the expression:
$${\color{red}{\int{\frac{x^{5} - 1}{x^{3}} d x}}} = {\color{red}{\int{\left(x^{2} - \frac{1}{x^{3}}\right)d x}}}$$
Gliedweise integrieren:
$${\color{red}{\int{\left(x^{2} - \frac{1}{x^{3}}\right)d x}}} = {\color{red}{\left(- \int{\frac{1}{x^{3}} d x} + \int{x^{2} d x}\right)}}$$
Wenden Sie die Potenzregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=2$$$ an:
$$- \int{\frac{1}{x^{3}} d x} + {\color{red}{\int{x^{2} d x}}}=- \int{\frac{1}{x^{3}} d x} + {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=- \int{\frac{1}{x^{3}} d x} + {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$
Wenden Sie die Potenzregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=-3$$$ an:
$$\frac{x^{3}}{3} - {\color{red}{\int{\frac{1}{x^{3}} d x}}}=\frac{x^{3}}{3} - {\color{red}{\int{x^{-3} d x}}}=\frac{x^{3}}{3} - {\color{red}{\frac{x^{-3 + 1}}{-3 + 1}}}=\frac{x^{3}}{3} - {\color{red}{\left(- \frac{x^{-2}}{2}\right)}}=\frac{x^{3}}{3} - {\color{red}{\left(- \frac{1}{2 x^{2}}\right)}}$$
Daher,
$$\int{\frac{x^{5} - 1}{x^{3}} d x} = \frac{x^{3}}{3} + \frac{1}{2 x^{2}}$$
Vereinfachen:
$$\int{\frac{x^{5} - 1}{x^{3}} d x} = \frac{2 x^{5} + 3}{6 x^{2}}$$
Fügen Sie die Integrationskonstante hinzu:
$$\int{\frac{x^{5} - 1}{x^{3}} d x} = \frac{2 x^{5} + 3}{6 x^{2}}+C$$
Antwort
$$$\int \frac{x^{5} - 1}{x^{3}}\, dx = \frac{2 x^{5} + 3}{6 x^{2}} + C$$$A