integral of (3-2x)/(5x+7)

Lösung

\int \frac{3 - 2 x}{5 x + 7} d x

- \frac{2}{5} x + \frac{29}{25} ln ∣ 5 x + 7 ∣ + C

Schritte zur Lösung

\int \frac{3 - 2 x}{5 x + 7} d x

Multipliziere aus

\frac{3 - 2 x}{5 x + 7} : \frac{3}{5 x + 7} - \frac{2 x}{5 x + 7}

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}{5 x + 7} d x - \int \frac{2 x}{5 x + 7} d x

\int \frac{3}{5 x + 7} d x = \frac{3}{5} ln ∣ 5 x + 7 ∣

\int \frac{2 x}{5 x + 7} d x = 2 (\frac{1}{5} x - \frac{7}{25} ln ∣ 5 x + 7 ∣)

= \frac{3}{5} ln ∣ 5 x + 7 ∣ - 2 (\frac{1}{5} x - \frac{7}{25} ln ∣ 5 x + 7 ∣)

Vereinfache

\frac{3}{5} ln ∣ 5 x + 7 ∣ - 2 (\frac{1}{5} x - \frac{7}{25} ln ∣ 5 x + 7 ∣) : - \frac{2}{5} x + \frac{29}{25} ln ∣ 5 x + 7 ∣

= - \frac{2}{5} x + \frac{29}{25} ln ∣ 5 x + 7 ∣

Füge eine Konstante zur Lösung hinzu

= - \frac{2}{5} x + \frac{29}{25} ln ∣ 5 x + 7 ∣ + C

\int_{1}^{2} 9 r^{4} ln (r) d r

y^{^{'}} + 2 t y = t

d er i v a t i v e \frac{cos ( x )}{x ^{8}}

t an g e n t y = 8 x sin (x)

\int_{0}^{0.5} 2 x d x