integral from 1 to 4 of 1/((1+sqrt(x)))

Lösung

\int_{1}^{4} \frac{1}{( 1 + x )} d x

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

Dezimale

1.18906 \dots

Schritte zur Lösung

\int_{1}^{4} \frac{1}{1 + x} d x

Wende U-Substitution an

= \int_{2}^{3} \frac{2 ( u - 1 )}{u} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{2}^{3} \frac{u - 1}{u} d u

Multipliziere aus

\frac{u - 1}{u} : 1 - \frac{1}{u}

= 2 \cdot \int_{2}^{3} 1 - \frac{1}{u} d u

Wende die Summenregel an:

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

= 2 (\int_{2}^{3} 1 d u - \int_{2}^{3} \frac{1}{u} d u)

\int_{2}^{3} 1 d u = 1

\int_{2}^{3} \frac{1}{u} d u = ln (3) - ln (2)

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

Vereinfache

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

\int_{0}^{2} 3 x^{2} - 2 x + 7 e^{x} d x

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

\int_{1}^{5} 5 d x

\int_{- 2}^{3} x^{2} - 1 d x

\int_{0}^{\frac{π}{3}} 3 cos (x) + 1 d x