integral of 1/(sqrt(z+4)-2)

Lösung

\int \frac{1}{z + 4 - 2} d z

- 4 (\frac{1}{2} ln \frac{z + 4}{2} + 1 - \frac{1}{2} ln \frac{z + 4}{2} - 1 - \frac{1}{2} z + 4) + 2 ln ∣ z ∣ + C

Schritte zur Lösung

\int \frac{1}{z + 4 - 2} d z

\frac{1}{z + 4 - 2} = \frac{z + 4 + 2}{( z + 4 ) ^{2} - ( 2 ) ^{2}}

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

Multipliziere aus

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

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

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

= - 4 (\frac{1}{2} ln \frac{z + 4}{2} + 1 - \frac{1}{2} ln \frac{z + 4}{2} - 1 - \frac{1}{2} z + 4) + 2 ln ∣ z ∣

Füge eine Konstante zur Lösung hinzu

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

\int \frac{81 - x ^{2}}{8 x ^{2}} d x

\int (x^{2} - π^{2})^{2} d x

\int x^{\frac{25}{2}} d x

\int \frac{2 x}{1 + 3 x ^{2}} d x

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