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

Lösung

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

\frac{( x + 1 ) ^{2}}{2} - 5 (x + 1) + 6 ln ∣ x + 1 ∣ + C

Schritte zur Lösung

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

Wende U-Substitution an

= \int \frac{u ^{2} - 5 u + 6}{u} d u

Multipliziere aus

\frac{u ^{2} - 5 u + 6}{u} : u - 5 + \frac{6}{u}

= \int u - 5 + \frac{6}{u} d u

Wende die Summenregel an:

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

= \int u d u - \int 5 d u + \int \frac{6}{u} d u

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

\int 5 d u = 5 u

\int \frac{6}{u} d u = 6 ln ∣ u ∣

= \frac{u ^{2}}{2} - 5 u + 6 ln ∣ u ∣

Setze in

u = x + 1

ein

= \frac{( x + 1 ) ^{2}}{2} - 5 (x + 1) + 6 ln ∣ x + 1 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{( x + 1 ) ^{2}}{2} - 5 (x + 1) + 6 ln ∣ x + 1 ∣ + C

\int_{- \infty}^{0} \frac{1}{3 - 7 x} d x

\int sin (2 x) sin (x) d x

\int sec (x) - tan (x) d x

in v erse l a pl a ce \frac{8.2}{4.2 x ^{2} + 4 x + 5}

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