integral 1/(x(x+1)^{1/2)}

Lösung

integral

\frac{1}{x ( x + 1 ) ^{\frac{1}{2}}}

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

Schritte zur Lösung

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

Wende U-Substitution an

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

Faktorisiere

u^{2} - 1 : - (- u^{2} + 1)

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

Entferne die Konstante:

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

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

Nutze das gemeinsame Integral :

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

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

Setze in

u = (x + 1)^{\frac{1}{2}}

ein

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

Vereinfache

2 - \frac{ln ( x + 1 ) ^{\frac{1}{2}} + 1}{2} - \frac{ln ( x + 1 ) ^{\frac{1}{2}} - 1}{2} : - ln (x + 1)^{\frac{1}{2}} + 1 + ln (x + 1)^{\frac{1}{2}} - 1

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

Füge eine Konstante zur Lösung hinzu

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

in t e g r a l \frac{x ^{2}}{( x - 2 ) ( x ^{2} + 1 )}

in t e g r a l y = \frac{x ^{2} - cos ( x ) + 1}{x}

in t e g r a l \frac{0.001}{( 0.001 x + 1 ) ^{2}}

in t e g r a l x - 3

in t e g r a l \frac{( 1 - tan ^{2} ( x ) )}{sec ^{2} ( x )}