integral of ((x+1))/(7x+x^2)

Lösung

\int \frac{( x + 1 )}{7 x + x ^{2}} d x

\frac{1}{7} ln ∣ x ∣ + \frac{6}{7} ln ∣ x + 7 ∣ + C

Schritte zur Lösung

\int \frac{x + 1}{7 x + x ^{2}} d x

Ermittle den Partialbruch von

\frac{x + 1}{7 x + x ^{2}} : \frac{1}{7 x} + \frac{6}{7 ( x + 7 )}

= \int \frac{1}{7 x} + \frac{6}{7 ( x + 7 )} d x

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{1}{7 x} d x + \int \frac{6}{7 ( x + 7 )} d x

\int \frac{1}{7 x} d x = \frac{1}{7} ln ∣ x ∣

\int \frac{6}{7 ( x + 7 )} d x = \frac{6}{7} ln ∣ x + 7 ∣

= \frac{1}{7} ln ∣ x ∣ + \frac{6}{7} ln ∣ x + 7 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{7} ln ∣ x ∣ + \frac{6}{7} ln ∣ x + 7 ∣ + C

\frac{d N}{d t} = N (1 - 0.001 N)

\int \frac{1}{( 1 - y ) ( 1 + y )} d y

\frac{d}{d x} (\frac{1}{3} x^{3} + \frac{1}{2} x^{2} - 6 x + 8)

y^{^{'}} = x sin (x) + x cos (x)

l a pl a ce t r an s f or m cos (4 x)