integral ((6x+1))/((2x-3))

Lösung

integral

\frac{( 6 x + 1 )}{( 2 x - 3 )}

3 x - \frac{9}{2} + 5 ln ∣ 2 x - 3 ∣ + C

Schritte zur Lösung

\int \frac{6 x + 1}{2 x - 3} d x

Multipliziere aus

\frac{6 x + 1}{2 x - 3} : \frac{6 x}{2 x - 3} + \frac{1}{2 x - 3}

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{6 x}{2 x - 3} d x + \int \frac{1}{2 x - 3} d x

\int \frac{6 x}{2 x - 3} d x = 3 x - \frac{9}{2} + \frac{9}{2} ln ∣ 2 x - 3 ∣

\int \frac{1}{2 x - 3} d x = \frac{1}{2} ln ∣ 2 x - 3 ∣

= 3 x - \frac{9}{2} + \frac{9}{2} ln ∣ 2 x - 3 ∣ + \frac{1}{2} ln ∣ 2 x - 3 ∣

Addiere gleiche Elemente:

\frac{9}{2} ln ∣ 2 x - 3 ∣ + \frac{1}{2} ln ∣ 2 x - 3 ∣ = 5 ln ∣ 2 x - 3 ∣

= 3 x - \frac{9}{2} + 5 ln ∣ 2 x - 3 ∣

Füge eine Konstante zur Lösung hinzu

= 3 x - \frac{9}{2} + 5 ln ∣ 2 x - 3 ∣ + C

in t e g r a l \frac{9}{( 3 x - 4 ) ^{2}}

in t e g r a l y^{4} (1 - y)^{3}

in t e g r a l \frac{1}{( x ) ^{2}}

in t e g r a l \frac{1}{( x ^{2} + 2 x + 1 ) ^{2}}

in t e g r a l \frac{4}{( 1 + x ^{2} )}