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

Lösung

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

2 ln x + 1 + x + C

Schritte zur Lösung

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

Wende Exponentenregel an:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

x (x + 1) = x (x + 1)

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

Nutze das gemeinsame Integral :

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

= 2 ln u^{2} + 1 + u

Setze in

u = x

ein

= 2 ln (x)^{2} + 1 + x

(x)^{2} = x

= 2 ln x + 1 + x

Füge eine Konstante zur Lösung hinzu

= 2 ln x + 1 + x + C

t an g e n t f (x) = x - 1, (10, 3)

\int_{0}^{3} 4 x + 5 d x

a re a x = - 3, x = - 1, y = x^{3} + 8, y = 0

\frac{\partial}{\partial x} (y^{2} e^{- x})

\frac{\partial}{\partial x} ((A x^{c} + B y^{c})^{\frac{1}{c}})