integral from 0 to 1 of (16x)/(8x^2+2)

Lösung

\int_{0}^{1} \frac{16 x}{8 x ^{2} + 2} d x

ln (5)

Dezimale

1.60943 \dots

Schritte zur Lösung

\int_{0}^{1} \frac{16 x}{8 x ^{2} + 2} d x

Entferne die Konstante:

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

= 16 \cdot \int_{0}^{1} \frac{x}{8 x ^{2} + 2} d x

Wende U-Substitution an

= 16 \cdot \int_{2}^{10} \frac{1}{16 u} d u

Entferne die Konstante:

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

= 16 \cdot \frac{1}{16} \cdot \int_{2}^{10} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= 16 \cdot \frac{1}{16} [ln ∣ u ∣]_{2}^{10}

Vereinfache

16 \cdot \frac{1}{16} [ln ∣ u ∣]_{2}^{10} : [ln ∣ u ∣]_{2}^{10}

= [ln ∣ u ∣]_{2}^{10}

Berechne die Grenzen:

ln (10) - ln (2)

= ln (10) - ln (2)

Vereinfache

= ln (5)

\int tan^{3} (x) sec^{- \frac{1}{2}} (x) d x

\frac{\partial}{\partial t} (sin (t))

x^{2} ln (\frac{x}{y}) d y + x y d x = 0

\frac{\partial}{\partial x} (sin (x e^{x y}))

y^{^{''}} - y^{^{'}} + \frac{1}{4} y = 2 + e^{\frac{x}{2}}