integral of (4-x^2)/(2+x)

Lösung

\int \frac{4 - x ^{2}}{2 + x} d x

4 x - \frac{1}{2} (x + 2)^{2} + 8 + C

Schritte zur Lösung

\int \frac{4 - x ^{2}}{2 + x} d x

Multipliziere aus

\frac{4 - x ^{2}}{2 + x} : \frac{4}{2 + x} - \frac{x ^{2}}{2 + 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{4}{2 + x} d x - \int \frac{x ^{2}}{2 + x} d x

\int \frac{4}{2 + x} d x = 4 ln ∣ 2 + x ∣

\int \frac{x ^{2}}{2 + x} d x = \frac{( 2 + x ) ^{2}}{2} - 4 (2 + x) + 4 ln ∣ 2 + x ∣

= 4 ln ∣ 2 + x ∣ - (\frac{( 2 + x ) ^{2}}{2} - 4 (2 + x) + 4 ln ∣ 2 + x ∣)

Vereinfache

4 ln ∣ 2 + x ∣ - (\frac{( 2 + x ) ^{2}}{2} - 4 (2 + x) + 4 ln ∣ 2 + x ∣) : 4 x - \frac{1}{2} (x + 2)^{2} + 8

= 4 x - \frac{1}{2} (x + 2)^{2} + 8

Füge eine Konstante zur Lösung hinzu

= 4 x - \frac{1}{2} (x + 2)^{2} + 8 + C

\frac{d y}{d t} = 3 y

t an g e n t y = x^{3 - 5 x^{2}} + 7 x + 3

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

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

a re a - 3 x^{2} - 10 x, 3