integral of ((x+1))/(x^2+5x+6)

Lösung

\int \frac{( x + 1 )}{x ^{2} + 5 x + 6} d x

2 (\frac{1}{4} ln 4 x^{2} + 20 x + 24 + \frac{3}{4} (ln ∣ 2 x + 6 ∣ - ln ∣ 2 x + 4 ∣)) + C

Schritte zur Lösung

\int \frac{x + 1}{x ^{2} + 5 x + 6} d x

Vervollständige das Quadrat

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

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

Wende U-Substitution an

= \int \frac{2 ( 2 u - 3 )}{4 u ^{2} - 1} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{2 u - 3}{4 u ^{2} - 1} d u

Multipliziere aus

\frac{2 u - 3}{4 u ^{2} - 1} : \frac{2 u}{4 u ^{2} - 1} - \frac{3}{4 u ^{2} - 1}

Wende die Summenregel an:

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

= 2 (\int \frac{2 u}{4 u ^{2} - 1} d u - \int \frac{3}{4 u ^{2} - 1} d u)

\int \frac{2 u}{4 u ^{2} - 1} d u = \frac{1}{4} ln 4 u^{2} - 1

\int \frac{3}{4 u ^{2} - 1} d u = - \frac{3}{4} (ln ∣ 2 u + 1 ∣ - ln ∣ 2 u - 1 ∣)

= 2 (\frac{1}{4} ln 4 u^{2} - 1 - (- \frac{3}{4} (ln ∣ 2 u + 1 ∣ - ln ∣ 2 u - 1 ∣)))

Setze in

u = x + \frac{5}{2}

ein

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

Vereinfache

2 (\frac{1}{4} ln 4 (x + \frac{5}{2})^{2} - 1 - (- \frac{3}{4} (ln 2 (x + \frac{5}{2}) + 1 - ln 2 (x + \frac{5}{2}) - 1))) : 2 (\frac{1}{4} ln 4 x^{2} + 20 x + 24 + \frac{3}{4} (ln ∣ 2 x + 6 ∣ - ln ∣ 2 x + 4 ∣))

= 2 (\frac{1}{4} ln 4 x^{2} + 20 x + 24 + \frac{3}{4} (ln ∣ 2 x + 6 ∣ - ln ∣ 2 x + 4 ∣))

Füge eine Konstante zur Lösung hinzu

= 2 (\frac{1}{4} ln 4 x^{2} + 20 x + 24 + \frac{3}{4} (ln ∣ 2 x + 6 ∣ - ln ∣ 2 x + 4 ∣)) + C

d er i v a t i v e 1 - 3 x

\int \frac{tan ( ln ( 2 x + 4 ) )}{2 x + 4} d x

t a y l or 4 x + 1

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

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