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

Lösung

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

ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x + C

Schritte zur Lösung

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

Entferne die Konstante:

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

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

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

= 2 \cdot \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

= 2 (\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

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

Vereinfache

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

= ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x

Füge eine Konstante zur Lösung hinzu

= ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x + C

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

\int \frac{4 x ^{2}}{( x ^{2} + 1 ) ^{2}} d x

\int t sin (nπ t) d t

\int x^{2} + 3 x + 3 d x

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