integral of 5/(8x(x+1)^2)

Lösung

\int \frac{5}{8 x ( x + 1 ) ^{2}} d x

\frac{5 ( ln ∣ x ∣ ( x + 1 ) - ln ∣ x + 1 ∣ ( x + 1 ) + 1 )}{8 ( x + 1 )} + C

Schritte zur Lösung

\int \frac{5}{8 x ( x + 1 ) ^{2}} d x

Entferne die Konstante:

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

= \frac{5}{8} \cdot \int \frac{1}{x ( x + 1 ) ^{2}} d x

Wende U-Substitution an

= \frac{5}{8} \cdot \int \frac{1}{( u - 1 ) u ^{2}} d u

Ermittle den Partialbruch von

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

= \frac{5}{8} \cdot \int \frac{1}{u - 1} - \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{5}{8} (\int \frac{1}{u - 1} d u - \int \frac{1}{u} d u - \int \frac{1}{u ^{2}} d u)

\int \frac{1}{u - 1} d u = ln ∣ u - 1 ∣

\int \frac{1}{u} d u = ln ∣ u ∣

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

= \frac{5}{8} (ln ∣ u - 1 ∣ - ln ∣ u ∣ - (- \frac{1}{u}))

Setze in

u = x + 1

ein

= \frac{5}{8} (ln ∣ x + 1 - 1 ∣ - ln ∣ x + 1 ∣ - (- \frac{1}{x + 1}))

Vereinfache

\frac{5}{8} (ln ∣ x + 1 - 1 ∣ - ln ∣ x + 1 ∣ - (- \frac{1}{x + 1})) : \frac{5 ( ln ∣ x ∣ ( x + 1 ) - ln ∣ x + 1 ∣ ( x + 1 ) + 1 )}{8 ( x + 1 )}

= \frac{5 ( ln ∣ x ∣ ( x + 1 ) - ln ∣ x + 1 ∣ ( x + 1 ) + 1 )}{8 ( x + 1 )}

Füge eine Konstante zur Lösung hinzu

= \frac{5 ( ln ∣ x ∣ ( x + 1 ) - ln ∣ x + 1 ∣ ( x + 1 ) + 1 )}{8 ( x + 1 )} + C

\int - e^{- 2 y} + 1 d y

t an g e n t y = \frac{x}{x - 5}, (- 1, a)

\int_{6}^{8} \frac{48}{( x - 6 ) ^{3}} d x

\frac{\partial}{\partial x} (\frac{x}{y + 8 z})

\int_{1}^{\infty} 6 e^{- 3 x} d x