integral of (2x^2-1)/(x(x+1)^3)

Lösung

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

- \frac{3}{x + 1} + \frac{1}{( x + 1 ) ^{2}} - ln ∣ x ∣ + ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 ) ^{2}} + C

Schritte zur Lösung

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

Multipliziere aus

\frac{2 x ^{2} - 1}{x ( x + 1 ) ^{3}} : \frac{2 x}{( x + 1 ) ^{3}} - \frac{1}{x ( x + 1 ) ^{3}}

= \int \frac{2 x}{( x + 1 ) ^{3}} - \frac{1}{x ( x + 1 ) ^{3}} d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{2 x}{( x + 1 ) ^{3}} d x - \int \frac{1}{x ( x + 1 ) ^{3}} d x

\int \frac{2 x}{( x + 1 ) ^{3}} d x = - \frac{2}{x + 1} + \frac{1}{( x + 1 ) ^{2}}

\int \frac{1}{x ( x + 1 ) ^{3}} d x = ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{x + 1} + \frac{1}{2 ( x + 1 ) ^{2}}

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

Vereinfache

- \frac{2}{x + 1} + \frac{1}{( x + 1 ) ^{2}} - (ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{x + 1} + \frac{1}{2 ( x + 1 ) ^{2}}) : - \frac{3}{x + 1} + \frac{1}{( x + 1 ) ^{2}} - ln ∣ x ∣ + ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 ) ^{2}}

= - \frac{3}{x + 1} + \frac{1}{( x + 1 ) ^{2}} - ln ∣ x ∣ + ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 ) ^{2}}

Füge eine Konstante zur Lösung hinzu

= - \frac{3}{x + 1} + \frac{1}{( x + 1 ) ^{2}} - ln ∣ x ∣ + ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 ) ^{2}} + C

\frac{d}{d x} (tan (1 - x))

\frac{\partial}{\partial x} (x^{2} + 10 x y + y^{2})

x \to 0 + lim (x \cdot (e^{\frac{1}{x}} - 1))

\int (\frac{1}{x} + \frac{x x}{3}) d x

\frac{d}{d x} (arctan (x))