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

Lösung

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

2 (\frac{1}{4} ln 4 x^{2} + 12 x - 16 - \frac{1}{20} (ln \frac{8}{5} + \frac{2}{5} x - ln - \frac{2}{5} + \frac{2}{5} x)) + C

Schritte zur Lösung

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

Vervollständige das Quadrat

x^{2} + 3 x - 4 : (x + \frac{3}{2})^{2} - \frac{25}{4}

= \int \frac{x + 2}{( x + \frac{3}{2} ) ^{2} - \frac{25}{4}} d x

Wende U-Substitution an

= \int \frac{2 ( 2 u + 1 )}{4 u ^{2} - 25} 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 + 1}{4 u ^{2} - 25} d u

Multipliziere aus

\frac{2 u + 1}{4 u ^{2} - 25} : \frac{2 u}{4 u ^{2} - 25} + \frac{1}{4 u ^{2} - 25}

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} - 25} d u + \int \frac{1}{4 u ^{2} - 25} d u)

\int \frac{2 u}{4 u ^{2} - 25} d u = \frac{1}{4} ln 4 u^{2} - 25

\int \frac{1}{4 u ^{2} - 25} d u = - \frac{1}{20} (ln \frac{2}{5} u + 1 - ln \frac{2}{5} u - 1)

= 2 (\frac{1}{4} ln 4 u^{2} - 25 - \frac{1}{20} (ln \frac{2}{5} u + 1 - ln \frac{2}{5} u - 1))

Setze in

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

ein

= 2 (\frac{1}{4} ln 4 (x + \frac{3}{2})^{2} - 25 - \frac{1}{20} (ln \frac{2}{5} (x + \frac{3}{2}) + 1 - ln \frac{2}{5} (x + \frac{3}{2}) - 1))

Vereinfache

2 (\frac{1}{4} ln 4 (x + \frac{3}{2})^{2} - 25 - \frac{1}{20} (ln \frac{2}{5} (x + \frac{3}{2}) + 1 - ln \frac{2}{5} (x + \frac{3}{2}) - 1)) : 2 (\frac{1}{4} ln 4 x^{2} + 12 x - 16 - \frac{1}{20} (ln \frac{8}{5} + \frac{2}{5} x - ln - \frac{2}{5} + \frac{2}{5} x))

= 2 (\frac{1}{4} ln 4 x^{2} + 12 x - 16 - \frac{1}{20} (ln \frac{8}{5} + \frac{2}{5} x - ln - \frac{2}{5} + \frac{2}{5} x))

Füge eine Konstante zur Lösung hinzu

= 2 (\frac{1}{4} ln 4 x^{2} + 12 x - 16 - \frac{1}{20} (ln \frac{8}{5} + \frac{2}{5} x - ln - \frac{2}{5} + \frac{2}{5} x)) + C

\int \frac{( e ^{x} - 3 )}{x} d x

\int cos (n t) d t

\frac{d}{d x} (x^{3} (x - 4))

d er i v a t i v e 91 - 90 e^{- 0.2 t}

in v erse l a pl a ce \frac{6}{s} + 16 \cdot \frac{1}{s ^{2} + 16}