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

Lösung

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

\frac{1}{7} ln ∣ 7 x - 2 ∣ - \frac{1}{7} ln ∣ 7 x + 2 ∣ + C

Schritte zur Lösung

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

\int \frac{1}{7 x - 2} d x = \frac{1}{7} ln ∣ 7 x - 2 ∣

\int \frac{1}{7 x + 2} d x = \frac{1}{7} ln ∣ 7 x + 2 ∣

= \frac{1}{7} ln ∣ 7 x - 2 ∣ - \frac{1}{7} ln ∣ 7 x + 2 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{7} ln ∣ 7 x - 2 ∣ - \frac{1}{7} ln ∣ 7 x + 2 ∣ + C

\int (\frac{2 + x + x}{x}) d x

\int \frac{3 x + 2}{x ^{2} - 3 x + 2} d x

\int \frac{e ^{t}}{e ^{t} - 2} d t

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\int \frac{16 x ^{3}}{4 x ^{2} - 4 x + 1} d x