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

Lösung

integral

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

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

Schritte zur Lösung

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

Lange Division

\frac{( x + 1 ) ^{2}}{( x - 1 ) ^{2}} : 1 + \frac{4 x}{x ^{2} - 2 x + 1}

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

\int 1 d x = x

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

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

Füge eine Konstante zur Lösung hinzu

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

in t e g r a l tan (2 - 3 x)

in t e g r a l e^{x^{2}} x^{2}

in t e g r a l 3 x^{2} - 9 x + 5

in t e g r a l \frac{e ^{x}}{3}

in t e g r a l \frac{1}{5 x}