integral of (x^4-x^2+1/(4x^3)-1/(x^2))

Lösung

\int (x^{4} - x^{2} + \frac{1}{4 x ^{3}} - \frac{1}{x ^{2}}) d x

\frac{x ^{5}}{5} - \frac{x ^{3}}{3} - \frac{1}{8 x ^{2}} + \frac{1}{x} + C

Schritte zur Lösung

\int x^{4} - x^{2} + \frac{1}{4 x ^{3}} - \frac{1}{x ^{2}} d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x^{4} d x - \int x^{2} d x + \int \frac{1}{4 x ^{3}} d x - \int \frac{1}{x ^{2}} d x

\int x^{4} d x = \frac{x ^{5}}{5}

\int x^{2} d x = \frac{x ^{3}}{3}

\int \frac{1}{4 x ^{3}} d x = - \frac{1}{8 x ^{2}}

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

= \frac{x ^{5}}{5} - \frac{x ^{3}}{3} - \frac{1}{8 x ^{2}} - (- \frac{1}{x})

Vereinfache

= \frac{x ^{5}}{5} - \frac{x ^{3}}{3} - \frac{1}{8 x ^{2}} + \frac{1}{x}

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{5}}{5} - \frac{x ^{3}}{3} - \frac{1}{8 x ^{2}} + \frac{1}{x} + C