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

Lösung

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

\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2} - x + C

Schritte zur Lösung

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

Multipliziere aus

(1 - x) (1 + x) : 1 - x^{2}

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

\frac{x ^{2}}{1 - x ^{2}} = \frac{1}{1 - x ^{2}} - 1

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

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

\int 1 d x = x

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

Füge eine Konstante zur Lösung hinzu

= \frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2} - x + C

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

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

\int (x + 2) e^{2 x} d x

\int sin (u) e^{u} d u

\int \frac{3}{sin ( 3 x )} d x