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

Lösung

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

\frac{2}{3} (x - 1)^{\frac{3}{2}} - x + 1 + 4 x - 1 - 4 ln x - 1 + 1 + C

Schritte zur Lösung

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

Wende U-Substitution an

= \int \frac{u + 1}{u + 1} d u

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Lange Division

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

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

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 v^{2} d v - \int v d v + \int 2 d v - \int \frac{2}{v + 1} d v)

\int v^{2} d v = \frac{v ^{3}}{3}

\int v d v = \frac{v ^{2}}{2}

\int 2 d v = 2 v

\int \frac{2}{v + 1} d v = 2 ln ∣ v + 1 ∣

= 2 (\frac{v ^{3}}{3} - \frac{v ^{2}}{2} + 2 v - 2 ln ∣ v + 1 ∣)

Ersetze zurück

= 2 (\frac{( x - 1 ) ^{3}}{3} - \frac{( x - 1 ) ^{2}}{2} + 2 x - 1 - 2 ln x - 1 + 1)

Vereinfache

2 (\frac{( x - 1 ) ^{3}}{3} - \frac{( x - 1 ) ^{2}}{2} + 2 x - 1 - 2 ln x - 1 + 1) : \frac{2}{3} (x - 1)^{\frac{3}{2}} - x + 1 + 4 x - 1 - 4 ln x - 1 + 1

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

Füge eine Konstante zur Lösung hinzu

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

f (x) = 2 e^{- x} x - e^{- x} x^{2}

d er i v a t i v e \frac{x ^{2}}{x + 5}

a re a 1 - 2 x^{2}, ∣ x ∣

t a y l or \frac{1}{1 - x}, 7

\int (x^{7} - 5 x^{- 2} + 4) d x