integral of (3z+1)/((z^2-4)^2)

Lösung

\int \frac{3 z + 1}{( z ^{2} - 4 ) ^{2}} d z

- \frac{3}{2 ( z ^{2} - 4 )} + \frac{1}{32} ln ∣ z + 2 ∣ - \frac{1}{16 ( z + 2 )} - \frac{1}{32} ln ∣ z - 2 ∣ - \frac{1}{16 ( z - 2 )} + C

Schritte zur Lösung

\int \frac{3 z + 1}{( z ^{2} - 4 ) ^{2}} d z

Multipliziere aus

\frac{3 z + 1}{( z ^{2} - 4 ) ^{2}} : \frac{3 z}{( z ^{2} - 4 ) ^{2}} + \frac{1}{( z ^{2} - 4 ) ^{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 z}{( z ^{2} - 4 ) ^{2}} d z + \int \frac{1}{( z ^{2} - 4 ) ^{2}} d z

\int \frac{3 z}{( z ^{2} - 4 ) ^{2}} d z = - \frac{3}{2 ( z ^{2} - 4 )}

\int \frac{1}{( z ^{2} - 4 ) ^{2}} d z = \frac{1}{32} ln ∣ z + 2 ∣ - \frac{1}{16 ( z + 2 )} - \frac{1}{32} ln ∣ z - 2 ∣ - \frac{1}{16 ( z - 2 )}

= - \frac{3}{2 ( z ^{2} - 4 )} + \frac{1}{32} ln ∣ z + 2 ∣ - \frac{1}{16 ( z + 2 )} - \frac{1}{32} ln ∣ z - 2 ∣ - \frac{1}{16 ( z - 2 )}

Füge eine Konstante zur Lösung hinzu

= - \frac{3}{2 ( z ^{2} - 4 )} + \frac{1}{32} ln ∣ z + 2 ∣ - \frac{1}{16 ( z + 2 )} - \frac{1}{32} ln ∣ z - 2 ∣ - \frac{1}{16 ( z - 2 )} + C

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