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

Lösung

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

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

Schritte zur Lösung

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

Wende U-Substitution an

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

(u^{2} - 1)^{2} = u^{4} - 2 u^{2} + 1

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

Ermittle den Partialbruch von

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

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

\int \frac{1}{4 ( u + 1 )} d u = \frac{1}{4} ln ∣ u + 1 ∣

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

\int \frac{1}{4 ( u - 1 )} d u = \frac{1}{4} ln ∣ u - 1 ∣

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

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

Setze in

u = x + 1

ein

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

Füge eine Konstante zur Lösung hinzu

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

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