integral of x/((1+2x)^2)

Lösung

\int \frac{x}{( 1 + 2 x ) ^{2}} d x

\frac{1}{4} (ln ∣ 1 + 2 x ∣ + \frac{1}{1 + 2 x}) + C

Schritte zur Lösung

\int \frac{x}{( 1 + 2 x ) ^{2}} d x

Wende U-Substitution an

= \int \frac{u - 1}{4 u ^{2}} d u

Entferne die Konstante:

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

= \frac{1}{4} \cdot \int \frac{u - 1}{u ^{2}} d u

Multipliziere aus

\frac{u - 1}{u ^{2}} : \frac{1}{u} - \frac{1}{u ^{2}}

= \frac{1}{4} \cdot \int \frac{1}{u} - \frac{1}{u ^{2}} 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

= \frac{1}{4} (\int \frac{1}{u} d u - \int \frac{1}{u ^{2}} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

= \frac{1}{4} (ln ∣ u ∣ - (- \frac{1}{u}))

Setze in

u = 1 + 2 x

ein

= \frac{1}{4} (ln ∣ 1 + 2 x ∣ - (- \frac{1}{1 + 2 x}))

Vereinfache

= \frac{1}{4} (ln ∣ 1 + 2 x ∣ + \frac{1}{1 + 2 x})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4} (ln ∣ 1 + 2 x ∣ + \frac{1}{1 + 2 x}) + C

\frac{\partial}{\partial t} (2 s^{2} t^{2})

\int x^{2} 8 + x d x

\int \frac{7}{9 + 4 x ^{2}} d x

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

y^{^{''}} + 5 y^{^{'}} + 6 = 1