integral from 0 to 1 of 1/(sqrt(x+x^2))

Lösung

\int_{0}^{1} \frac{1}{x + x ^{2}} d x

2 ln (1 + 2)

Dezimale

1.76274 \dots

Schritte zur Lösung

\int_{0}^{1} \frac{1}{x + x ^{2}} d x

Faktorisiere

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

= \int_{0}^{1} \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_{0}^{1} \frac{1}{x x + 1} d x

Wende U-Substitution an

= \int_{0}^{1} \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_{0}^{1} \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]_{0}^{1}

Berechne die Grenzen:

ln (1 + 2)

= 2 ln (1 + 2)

\int_{0}^{1} 4 t^{3} d t

\int_{0}^{a} a^{2} x - x^{3} d x

\int_{0}^{x} 2 t cos (t) d t

\int_{- π}^{π} a sin (θ) d θ

\int_{0}^{6} x^{2} - 9 d x