integral of ((x^2-1))/(x-1)

Lösung

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

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

Schritte zur Lösung

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

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

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

Vereinfache

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

= \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

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