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

Lösung

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

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

Schritte zur Lösung

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

Multipliziere aus

\frac{3 + z}{1 - z ^{2}} : \frac{3}{1 - z ^{2}} + \frac{z}{1 - z ^{2}}

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

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

\int \frac{z}{1 - z ^{2}} d z = - \frac{1}{2} ln 1 - z^{2}

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

Füge eine Konstante zur Lösung hinzu

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

n = 0 \sum \infty \frac{1}{n ^{2}} x^{n}

a re a y = \frac{2}{3} x^{\frac{3}{2}} + 3, [0, 2]

d er i v a t i v e y = x^{n}

\frac{6}{y} d x = (e^{3 x} + 1) d y

f (x) = 1 + x^{2}