integral of 1/(x^2+7x+12)

Lösung

\int \frac{1}{x ^{2} + 7 x + 12} d x

ln ∣ x + 3 ∣ - ln ∣ x + 4 ∣ + C

Schritte zur Lösung

\int \frac{1}{x ^{2} + 7 x + 12} d x

Ermittle den Partialbruch von

\frac{1}{x ^{2} + 7 x + 12} : \frac{1}{x + 3} - \frac{1}{x + 4}

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

\int \frac{1}{x + 3} d x = ln ∣ x + 3 ∣

\int \frac{1}{x + 4} d x = ln ∣ x + 4 ∣

= ln ∣ x + 3 ∣ - ln ∣ x + 4 ∣

Füge eine Konstante zur Lösung hinzu

= ln ∣ x + 3 ∣ - ln ∣ x + 4 ∣ + C

\int (sec (x) tan (x) - (sec (x))^{2}) d x

\frac{d}{d x} (\frac{[ 3 x ^{2} ]}{[ x ^{2} - 4 ]})

d er i v a t i v e f (x) = 3 x^{\frac{1}{3}} - 5 x^{\frac{4}{3}}

in v erse l a pl a ce \frac{1}{( ( s - 1 ) ^{2} + 4 )}

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