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

Lösung

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

4 (\frac{1}{2 2} x ln \frac{1}{2} (x + 2) - 1 - \frac{1}{2 2} x ln \frac{1}{2} x + 2 - 1 + \frac{2 + 1}{2} ln \frac{1}{2} (x + 2) + 1 + \frac{1 - 2}{2} ln \frac{x}{2} + 2 - 1 - 1) - \frac{5}{2 2} (ln \frac{x + 2}{2} + 1 - ln \frac{x + 2}{2} - 1) + C

Schritte zur Lösung

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

Multipliziere aus

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

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

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

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

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

Füge eine Konstante zur Lösung hinzu

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

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

\frac{d y}{d x} = (\frac{( 2 y + 3 )}{( 4 x + 5 )})^{2}

a re a cos (6 x), 0

y^{^{''}} + 43 y^{^{'}} + 12 y = 0, y (0) = 1, y^{^{'}} (0) = 0

n = 1 \sum \infty \frac{2}{n + 1}