integral ((3t+4))/((2t+3))

Lösung

integral

\frac{( 3 t + 4 )}{( 2 t + 3 )}

\frac{3}{2} t + \frac{9}{4} - \frac{1}{4} ln ∣ 2 t + 3 ∣ + C

Schritte zur Lösung

\int \frac{3 t + 4}{2 t + 3} d t

Multipliziere aus

\frac{3 t + 4}{2 t + 3} : \frac{3 t}{2 t + 3} + \frac{4}{2 t + 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{3 t}{2 t + 3} d t + \int \frac{4}{2 t + 3} d t

\int \frac{3 t}{2 t + 3} d t = \frac{3}{4} (2 t + 3 - 3 ln ∣ 2 t + 3 ∣)

\int \frac{4}{2 t + 3} d t = 2 ln ∣ 2 t + 3 ∣

= \frac{3}{4} (2 t + 3 - 3 ln ∣ 2 t + 3 ∣) + 2 ln ∣ 2 t + 3 ∣

Vereinfache

\frac{3}{4} (2 t + 3 - 3 ln ∣ 2 t + 3 ∣) + 2 ln ∣ 2 t + 3 ∣ : \frac{3}{2} t + \frac{9}{4} - \frac{1}{4} ln ∣ 2 t + 3 ∣

= \frac{3}{2} t + \frac{9}{4} - \frac{1}{4} ln ∣ 2 t + 3 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{3}{2} t + \frac{9}{4} - \frac{1}{4} ln ∣ 2 t + 3 ∣ + C

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

in t e g r a l (4 x + 9)^{\frac{1}{2}}

in t e g r a l (2 x + 8)^{3}

in t e g r a l \frac{1}{1 + x ^{3}}

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