integral of 1/x*sqrt((x-9)/(x+9))

Lösung

\int \frac{1}{x} \cdot \frac{x - 9}{x + 9} d x

ln \frac{x - 9}{x + 9} + 1 - ln \frac{x - 9}{x + 9} - 1 - 2 arctan (\frac{x - 9}{x + 9}) + C

Schritte zur Lösung

\int \frac{1}{x} \frac{x - 9}{x + 9} d x

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Ermittle den Partialbruch von

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

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

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

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

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

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

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

Setze in

u = \frac{x - 9}{x + 9}

ein

= 4 (\frac{1}{4} ln \frac{x - 9}{x + 9} + 1 - \frac{1}{4} ln \frac{x - 9}{x + 9} - 1 - \frac{1}{2} arctan (\frac{x - 9}{x + 9}))

Vereinfache

4 (\frac{1}{4} ln \frac{x - 9}{x + 9} + 1 - \frac{1}{4} ln \frac{x - 9}{x + 9} - 1 - \frac{1}{2} arctan (\frac{x - 9}{x + 9})) : ln \frac{x - 9}{x + 9} + 1 - ln \frac{x - 9}{x + 9} - 1 - 2 arctan (\frac{x - 9}{x + 9})

= ln \frac{x - 9}{x + 9} + 1 - ln \frac{x - 9}{x + 9} - 1 - 2 arctan (\frac{x - 9}{x + 9})

Füge eine Konstante zur Lösung hinzu

= ln \frac{x - 9}{x + 9} + 1 - ln \frac{x - 9}{x + 9} - 1 - 2 arctan (\frac{x - 9}{x + 9}) + C

\int x^{3} (3 x^{2} - 4)^{\frac{5}{2}} d x

\int 4 + 10 (x + 5)^{- 2} d x

\int 4 x + 9 d x

\int \frac{5 x ^{4}}{( x ^{5} + 1 ) ^{2}} d x

\int \frac{5}{x ^{2} + 2 x + 5} d x