integral 1/((t^2+t))

Lösung

integral

\frac{1}{( t ^{2} + t )}

ln ∣ t ∣ - ln ∣ t + 1 ∣ + C

Schritte zur Lösung

\int \frac{1}{t ^{2} + t} d t

Ermittle den Partialbruch von

\frac{1}{t ^{2} + t} : \frac{1}{t} - \frac{1}{t + 1}

= \int \frac{1}{t} - \frac{1}{t + 1} d t

Wende die Summenregel an:

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

= \int \frac{1}{t} d t - \int \frac{1}{t + 1} d t

\int \frac{1}{t} d t = ln ∣ t ∣

\int \frac{1}{t + 1} d t = ln ∣ t + 1 ∣

= ln ∣ t ∣ - ln ∣ t + 1 ∣

Füge eine Konstante zur Lösung hinzu

= ln ∣ t ∣ - ln ∣ t + 1 ∣ + C

in t e g r a l \frac{( 2 x + 1 )}{( x + 2 )}

in t e g r a l \frac{1}{x ( x ^{8} - 1 )}

in t e g r a l \frac{( 8 x + 2 )}{x}

in t e g r a l \frac{5}{x ^{4}}

in t e g r a l \frac{( 3 x - 2 )}{( 2 x - 1 )}