integral from 1 to 4 of 1/(1+2sqrt(x))

Lösung

\int_{1}^{4} \frac{1}{1 + 2 x} d x

\frac{- ln ( 5 ) + ln ( 3 )}{2} + 1

Dezimale

0.74458 \dots

Schritte zur Lösung

\int_{1}^{4} \frac{1}{1 + 2 x} d x

Wende U-Substitution an

= \int_{1}^{2} \frac{2 u}{1 + 2 u} d u

Entferne die Konstante:

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

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

Wende U-Substitution an

= 2 \cdot \int_{3}^{5} \frac{v - 1}{4 v} d v

Entferne die Konstante:

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

= 2 \cdot \frac{1}{4} \cdot \int_{3}^{5} \frac{v - 1}{v} d v

Multipliziere aus

\frac{v - 1}{v} : 1 - \frac{1}{v}

= 2 \cdot \frac{1}{4} \cdot \int_{3}^{5} 1 - \frac{1}{v} d v

Wende die Summenregel an:

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

= 2 \cdot \frac{1}{4} (\int_{3}^{5} 1 d v - \int_{3}^{5} \frac{1}{v} d v)

\int_{3}^{5} 1 d v = 2

\int_{3}^{5} \frac{1}{v} d v = ln (5) - ln (3)

= 2 \cdot \frac{1}{4} (2 - (ln (5) - ln (3)))

Vereinfache

2 \cdot \frac{1}{4} (2 - (ln (5) - ln (3))) : \frac{- ln ( 5 ) + ln ( 3 )}{2} + 1

= \frac{- ln ( 5 ) + ln ( 3 )}{2} + 1

\int_{1}^{\infty} \frac{7}{1 + x ^{2}} d x

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

\int_{0}^{2} 1 4^{2 x} d x

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

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