integral of ((2z-1))/(2z^2-6z+5)

Lösung

\int \frac{( 2 z - 1 )}{2 z ^{2} - 6 z + 5} d z

\frac{1}{2} ln 4 z^{2} - 12 z + 10 + 2 arctan (2 z - 3) + C

Schritte zur Lösung

\int \frac{2 z - 1}{2 z ^{2} - 6 z + 5} d z

Wende U-Substitution an

= \int \frac{u}{u ^{2} - 4 u + 5} d u

Vervollständige das Quadrat

u^{2} - 4 u + 5 : (u - 2)^{2} + 1

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

Wende U-Substitution an

= \int \frac{v + 2}{v ^{2} + 1} d v

Multipliziere aus

\frac{v + 2}{v ^{2} + 1} : \frac{v}{v ^{2} + 1} + \frac{2}{v ^{2} + 1}

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{v}{v ^{2} + 1} d v + \int \frac{2}{v ^{2} + 1} d v

\int \frac{v}{v ^{2} + 1} d v = \frac{1}{2} ln v^{2} + 1

\int \frac{2}{v ^{2} + 1} d v = 2 arctan (v)

= \frac{1}{2} ln v^{2} + 1 + 2 arctan (v)

Ersetze zurück

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

Vereinfache

\frac{1}{2} ln (2 z - 1 - 2)^{2} + 1 + 2 arctan (2 z - 1 - 2) : \frac{1}{2} ln 4 z^{2} - 12 z + 10 + 2 arctan (2 z - 3)

= \frac{1}{2} ln 4 z^{2} - 12 z + 10 + 2 arctan (2 z - 3)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln 4 z^{2} - 12 z + 10 + 2 arctan (2 z - 3) + C

(9 x + 1) y^{2} \frac{d y}{d x} + 4 x^{2} + 3 y^{3} = 0

\frac{d}{d x} (\frac{x ^{3.8}}{6 ^{x}})

d er i v a t i v e tan^{2} (6 θ)

in v erse l a pl a ce \frac{1}{s ^{3} ( s - 1 )}

\frac{\partial}{\partial x} (cos (2 y - 2 x))