integral of (-x+1)

Lösung

\int (- x + 1) d x

- \frac{x ^{2}}{2} + x + C

Schritte zur Lösung

\int - x + 1 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 d x + \int 1 d x

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

\int 1 d x = x

= - \frac{x ^{2}}{2} + x

Füge eine Konstante zur Lösung hinzu

= - \frac{x ^{2}}{2} + x + C

\int lo g_{10} (x^{2} + 1) d x

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

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

\int \frac{ln ( arctan ( x ) )}{1 + x ^{2}} d x

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