integral of (2x-3)/(sqrt(x^2+6x+4))

Lösung

\int \frac{2 x - 3}{x ^{2} + 6 x + 4} d x

2 x^{2} + 6 x + 4 - 9 ln \frac{1}{5} (x^{2} + 6 x + 4) + \frac{1}{5} (x + 3) + C

Schritte zur Lösung

\int \frac{2 x - 3}{x ^{2} + 6 x + 4} d x

Multipliziere aus

\frac{2 x - 3}{x ^{2} + 6 x + 4} : \frac{2 x}{x ^{2} + 6 x + 4} - \frac{3}{x ^{2} + 6 x + 4}

= \int \frac{2 x}{x ^{2} + 6 x + 4} - \frac{3}{x ^{2} + 6 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{2 x}{x ^{2} + 6 x + 4} d x - \int \frac{3}{x ^{2} + 6 x + 4} d x

\int \frac{2 x}{x ^{2} + 6 x + 4} d x = 2 (x^{2} + 6 x + 4 - 3 ln \frac{1}{5} (x^{2} + 6 x + 4) + \frac{1}{5} (x + 3))

\int \frac{3}{x ^{2} + 6 x + 4} d x = 3 ln \frac{1}{5} (x^{2} + 6 x + 4) + \frac{1}{5} (x + 3)

= 2 (x^{2} + 6 x + 4 - 3 ln \frac{1}{5} (x^{2} + 6 x + 4) + \frac{1}{5} (x + 3)) - 3 ln \frac{1}{5} (x^{2} + 6 x + 4) + \frac{1}{5} (x + 3)

Vereinfache

2 (x^{2} + 6 x + 4 - 3 ln \frac{1}{5} (x^{2} + 6 x + 4) + \frac{1}{5} (x + 3)) - 3 ln \frac{1}{5} (x^{2} + 6 x + 4) + \frac{1}{5} (x + 3) : 2 x^{2} + 6 x + 4 - 9 ln \frac{1}{5} (x^{2} + 6 x + 4) + \frac{1}{5} (x + 3)

= 2 x^{2} + 6 x + 4 - 9 ln \frac{1}{5} (x^{2} + 6 x + 4) + \frac{1}{5} (x + 3)

Füge eine Konstante zur Lösung hinzu

= 2 x^{2} + 6 x + 4 - 9 ln \frac{1}{5} (x^{2} + 6 x + 4) + \frac{1}{5} (x + 3) + C

\frac{d}{d x} (13 e^{x})

(x^{2} + 1) y^{^{'}} + 4 x (y - 1) = 0, y (0) = 3

\frac{d y}{d x} + y^{3} x + 8 y = 0

\int sin^{5} (9 x) d x

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