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

Lösung

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

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

Dezimale

1.18906 \dots

Schritte zur Lösung

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

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

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

\int_{0}^{3} \frac{x + 1}{1 + x + 1} 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_{0}^{2} 9 y^{2} y^{3} + 1 d y

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

\int_{0}^{3} (x - 3) e^{- x} d x

\int_{0}^{1} \frac{5 r}{( 4 + r ^{2} ) ^{2}} d r

\int_{1}^{1} 1