integral of x/(x^2-7x+12)

Lösung

\int \frac{x}{x ^{2} - 7 x + 12} d x

2 (\frac{1}{4} ln 4 x^{2} - 28 x + 48 - \frac{7}{4} (ln ∣ 2 x - 6 ∣ - ln ∣ 2 x - 8 ∣)) + C

Schritte zur Lösung

\int \frac{x}{x ^{2} - 7 x + 12} d x

Vervollständige das Quadrat

x^{2} - 7 x + 12 : (x - \frac{7}{2})^{2} - \frac{1}{4}

= \int \frac{x}{( x - \frac{7}{2} ) ^{2} - \frac{1}{4}} d x

Wende U-Substitution an

= \int \frac{2 ( 2 u + 7 )}{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 + 7}{4 u ^{2} - 1} d u

Multipliziere aus

\frac{2 u + 7}{4 u ^{2} - 1} : \frac{2 u}{4 u ^{2} - 1} + \frac{7}{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{7}{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{7}{4 u ^{2} - 1} d u = - \frac{7}{4} (ln ∣ 2 u + 1 ∣ - ln ∣ 2 u - 1 ∣)

= 2 (\frac{1}{4} ln 4 u^{2} - 1 - \frac{7}{4} (ln ∣ 2 u + 1 ∣ - ln ∣ 2 u - 1 ∣))

Setze in

u = x - \frac{7}{2}

ein

= 2 (\frac{1}{4} ln 4 (x - \frac{7}{2})^{2} - 1 - \frac{7}{4} (ln 2 (x - \frac{7}{2}) + 1 - ln 2 (x - \frac{7}{2}) - 1))

Vereinfache

2 (\frac{1}{4} ln 4 (x - \frac{7}{2})^{2} - 1 - \frac{7}{4} (ln 2 (x - \frac{7}{2}) + 1 - ln 2 (x - \frac{7}{2}) - 1)) : 2 (\frac{1}{4} ln 4 x^{2} - 28 x + 48 - \frac{7}{4} (ln ∣ 2 x - 6 ∣ - ln ∣ 2 x - 8 ∣))

= 2 (\frac{1}{4} ln 4 x^{2} - 28 x + 48 - \frac{7}{4} (ln ∣ 2 x - 6 ∣ - ln ∣ 2 x - 8 ∣))

Füge eine Konstante zur Lösung hinzu

= 2 (\frac{1}{4} ln 4 x^{2} - 28 x + 48 - \frac{7}{4} (ln ∣ 2 x - 6 ∣ - ln ∣ 2 x - 8 ∣)) + C

x \to \frac{1}{2} lim (- 4 x + 2)

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

\int_{0}^{\frac{π}{26}} tan (13 x) d x

5 x^{2} y^{^{'}} = y^{^{'}} + 2 x e^{- y}

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 7, y (0) = 7, y^{^{'}} (0) = 7