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Beliebt Rechnen >

integral of 1/(x^2sqrt(4+x))

Lösung

\int \frac{1}{x ^{2} 4 + x} d x

Lösung

2 (\frac{1}{32} ln 4 + x + 2 - \frac{1}{16 ( 4 + x + 2 )} - \frac{1}{32} ln 4 + x - 2 - \frac{1}{16 ( 4 + x - 2 )}) + C

Schritte zur Lösung

\int \frac{1}{x ^{2} 4 + x} d x

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Ermittle den Partialbruch von

\frac{1}{( u ^{2} - 4 ) ^{2}} : \frac{1}{32 ( u + 2 )} + \frac{1}{16 ( u + 2 ) ^{2}} - \frac{1}{32 ( u - 2 )} + \frac{1}{16 ( u - 2 ) ^{2}}

= 2 \cdot \int \frac{1}{32 ( u + 2 )} + \frac{1}{16 ( u + 2 ) ^{2}} - \frac{1}{32 ( u - 2 )} + \frac{1}{16 ( u - 2 ) ^{2}} d u

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{1}{32 ( u + 2 )} d u + \int \frac{1}{16 ( u + 2 ) ^{2}} d u - \int \frac{1}{32 ( u - 2 )} d u + \int \frac{1}{16 ( u - 2 ) ^{2}} d u)

\int \frac{1}{32 ( u + 2 )} d u = \frac{1}{32} ln ∣ u + 2 ∣

\int \frac{1}{16 ( u + 2 ) ^{2}} d u = - \frac{1}{16 ( u + 2 )}

\int \frac{1}{32 ( u - 2 )} d u = \frac{1}{32} ln ∣ u - 2 ∣

\int \frac{1}{16 ( u - 2 ) ^{2}} d u = - \frac{1}{16 ( u - 2 )}

= 2 (\frac{1}{32} ln ∣ u + 2 ∣ - \frac{1}{16 ( u + 2 )} - \frac{1}{32} ln ∣ u - 2 ∣ - \frac{1}{16 ( u - 2 )})

Setze in

u = 4 + x

ein

= 2 (\frac{1}{32} ln 4 + x + 2 - \frac{1}{16 ( 4 + x + 2 )} - \frac{1}{32} ln 4 + x - 2 - \frac{1}{16 ( 4 + x - 2 )})

Füge eine Konstante zur Lösung hinzu

= 2 (\frac{1}{32} ln 4 + x + 2 - \frac{1}{16 ( 4 + x + 2 )} - \frac{1}{32} ln 4 + x - 2 - \frac{1}{16 ( 4 + x - 2 )}) + C

Graph

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