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

Lösung

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

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

Schritte zur Lösung

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

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Lange Division

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

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

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

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

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

\int 2 d u = 2 u

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

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

Setze in

u = x

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

\frac{d y}{d x} - y = x y^{5}

\int 14 cos^{3} (x) d x

\int (2 x + 1) d x

s l o p e (5, - 1), (- 5, 5)

\frac{d}{d x} (3 x^{3} - 3 x + c)