integral of (5x+3)/(x^2+4x+4)

Lösung

\int \frac{5 x + 3}{x ^{2} + 4 x + 4} d x

5 ln ∣ x + 2 ∣ + \frac{7}{x + 2} + C

Schritte zur Lösung

\int \frac{5 x + 3}{x ^{2} + 4 x + 4} d x

Multipliziere aus

\frac{5 x + 3}{x ^{2} + 4 x + 4} : \frac{5 x}{x ^{2} + 4 x + 4} + \frac{3}{x ^{2} + 4 x + 4}

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{5 x}{x ^{2} + 4 x + 4} d x + \int \frac{3}{x ^{2} + 4 x + 4} d x

\int \frac{5 x}{x ^{2} + 4 x + 4} d x = 5 (ln ∣ x + 2 ∣ + \frac{2}{x + 2})

\int \frac{3}{x ^{2} + 4 x + 4} d x = - \frac{3}{x + 2}

= 5 (ln ∣ x + 2 ∣ + \frac{2}{x + 2}) - \frac{3}{x + 2}

Vereinfache

5 (ln ∣ x + 2 ∣ + \frac{2}{x + 2}) - \frac{3}{x + 2} : 5 ln ∣ x + 2 ∣ + \frac{7}{x + 2}

= 5 ln ∣ x + 2 ∣ + \frac{7}{x + 2}

Füge eine Konstante zur Lösung hinzu

= 5 ln ∣ x + 2 ∣ + \frac{7}{x + 2} + C

\frac{d}{d x} (2^{3^{x^{2}}})

\int x^{2} x^{3} + 29 d x

(1 - x^{2}) y^{^{'}} + x (y - 1) = - x^{3}

(3 x y + y^{2}) + (x^{2} + x y) y^{^{'}} = 0

\frac{d}{d x} (a^{x} - 4^{x})