integral of ((4x+13))/((x^2+3x-10))

Lösung

\int \frac{( 4 x + 13 )}{( x ^{2} + 3 x - 10 )} d x

2 ln 4 x^{2} + 12 x - 40 - ln \frac{10}{7} + \frac{2}{7} x + ln - \frac{4}{7} + \frac{2}{7} x + C

Schritte zur Lösung

\int \frac{4 x + 13}{x ^{2} + 3 x - 10} d x

Vervollständige das Quadrat

x^{2} + 3 x - 10 : (x + \frac{3}{2})^{2} - \frac{49}{4}

= \int \frac{4 x + 13}{( x + \frac{3}{2} ) ^{2} - \frac{49}{4}} d x

Wende U-Substitution an

= \int \frac{4 ( 4 u + 7 )}{4 u ^{2} - 49} d u

Entferne die Konstante:

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

= 4 \cdot \int \frac{4 u + 7}{4 u ^{2} - 49} d u

Multipliziere aus

\frac{4 u + 7}{4 u ^{2} - 49} : \frac{4 u}{4 u ^{2} - 49} + \frac{7}{4 u ^{2} - 49}

Wende die Summenregel an:

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

= 4 (\int \frac{4 u}{4 u ^{2} - 49} d u + \int \frac{7}{4 u ^{2} - 49} d u)

\int \frac{4 u}{4 u ^{2} - 49} d u = \frac{1}{2} ln 4 u^{2} - 49

\int \frac{7}{4 u ^{2} - 49} d u = - \frac{1}{4} (ln \frac{2}{7} u + 1 - ln \frac{2}{7} u - 1)

= 4 (\frac{1}{2} ln 4 u^{2} - 49 - \frac{1}{4} (ln \frac{2}{7} u + 1 - ln \frac{2}{7} u - 1))

Setze in

u = x + \frac{3}{2}

ein

= 4 (\frac{1}{2} ln 4 (x + \frac{3}{2})^{2} - 49 - \frac{1}{4} (ln \frac{2}{7} (x + \frac{3}{2}) + 1 - ln \frac{2}{7} (x + \frac{3}{2}) - 1))

Vereinfache

4 (\frac{1}{2} ln 4 (x + \frac{3}{2})^{2} - 49 - \frac{1}{4} (ln \frac{2}{7} (x + \frac{3}{2}) + 1 - ln \frac{2}{7} (x + \frac{3}{2}) - 1)) : 2 ln 4 x^{2} + 12 x - 40 - ln \frac{10}{7} + \frac{2}{7} x + ln - \frac{4}{7} + \frac{2}{7} x

= 2 ln 4 x^{2} + 12 x - 40 - ln \frac{10}{7} + \frac{2}{7} x + ln - \frac{4}{7} + \frac{2}{7} x

Füge eine Konstante zur Lösung hinzu

= 2 ln 4 x^{2} + 12 x - 40 - ln \frac{10}{7} + \frac{2}{7} x + ln - \frac{4}{7} + \frac{2}{7} x + C

t an g e n t 6 x^{2}, at x = 10

\int x^{2} - 4 x + 2 d x

\int sin (θ) \cdot cos (θ) d θ

\int 2 x - 2 d x

d er i v a t i v e x^{2} \cdot e^{x}