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

Lösung

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

ln b^{2} + 4 - ln (5)

Schritte zur Lösung

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

Entferne die Konstante:

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

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

Wende U-Substitution an

= 2 \cdot \int_{5}^{b^{2} + 4} \frac{1}{2 u} d u

Entferne die Konstante:

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

= 2 \cdot \frac{1}{2} \cdot \int_{5}^{b^{2} + 4} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} [ln ∣ u ∣]_{5}^{b^{2} + 4}

Vereinfache

2 \cdot \frac{1}{2} [ln ∣ u ∣]_{5}^{b^{2} + 4} : [ln ∣ u ∣]_{5}^{b^{2} + 4}

= [ln ∣ u ∣]_{5}^{b^{2} + 4}

Berechne die Grenzen:

ln b^{2} + 4 - ln (5)

= ln b^{2} + 4 - ln (5)

x^{2} y + y^{^{'}} = - 3 x^{2}

t an g e n t f (x) = 1 + 4 x, at x = 2

x \to c lim (x^{3} + 4 x^{2} - 3)

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

d er i v a t i v e \frac{- x ^{2} - 7}{( x ^{2} - 7 ) ^{2}}