integral from 0 to 4 of 1/(sqrt(2x+1))

Lösung

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

2

Schritte zur Lösung

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

Wende U-Substitution an

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

Integral einer Konstanten :

\int a d x = a x

= [1 \cdot u]_{1}^{3}

Vereinfache

= [u]_{1}^{3}

Berechne die Grenzen:

2

= 2

x \to - \infty lim (9 x^{3})

\frac{d}{d x} (\frac{m}{2 π ( a - x )})

\int_{0}^{\frac{π}{3}} 1

\frac{d}{d x} ((ln (x)) (ln (3 x - 7)))

\int \frac{ln ( x + 1 )}{x + 1} d x