integral from 0 to 3 of 1/(1+sqrt(1+x))

Lösung

\int_{0}^{3} \frac{1}{1 + 1 + x} d x

2 ln (2) + 2 - 2 ln (3)

Dezimale

1.18906 \dots

Schritte zur Lösung

\int_{0}^{3} \frac{1}{1 + 1 + x} d x

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

= \int_{0}^{3} - \frac{x + 1}{1 + 1 + x} + 1 d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{0}^{3} \frac{x + 1}{1 + 1 + x} d x + \int_{0}^{3} 1 d x

\int_{0}^{3} \frac{x + 1}{1 + 1 + x} d x = 2 ln (3) - 2 ln (2) + 1

\int_{0}^{3} 1 d x = 3

= - (2 ln (3) - 2 ln (2) + 1) + 3

Vereinfache

= 2 ln (2) + 2 - 2 ln (3)

\int_{- 1}^{2} (x) d x

\int_{- 4}^{4} 2 x^{3} - 2 x d x

\int_{2}^{4} x ln (x) d x

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

\int_{0}^{x} 4 e^{- 2 x} d x