integral from 0 to 8 of 1/(1+sqrt(1+x))

Lösung

\int_{0}^{8} \frac{1}{1 + 1 + x} d x

- 2 ln (2) + 4

Dezimale

2.61370 \dots

Schritte zur Lösung

\int_{0}^{8} \frac{1}{1 + 1 + x} d x

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

= \int_{0}^{8} - \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}^{8} \frac{x + 1}{1 + 1 + x} d x + \int_{0}^{8} 1 d x

\int_{0}^{8} \frac{x + 1}{1 + 1 + x} d x = 2 (ln (2) + 2)

\int_{0}^{8} 1 d x = 8

= - 2 (ln (2) + 2) + 8

Vereinfache

- 2 (ln (2) + 2) + 8 : - 2 ln (2) + 4

= - 2 ln (2) + 4

\int_{0}^{1} x^{3} (2 + x^{4})^{3} d x

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

\int_{0}^{1} 30 x^{4} (1 - x)^{2} d x

\int_{3}^{s i n (x + y)} e^{t^{3}} d t

\int_{1}^{\infty} (\frac{1}{3})^{x} d x