integral of 1/(2-t)+1/(2+t)

Lösung

\int \frac{1}{2 - t} + \frac{1}{2 + t} d t

- ln ∣ 2 - t ∣ + ln ∣ 2 + t ∣ + C

Schritte zur Lösung

\int \frac{1}{2 - t} + \frac{1}{2 + t} d t

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}{2 - t} d t + \int \frac{1}{2 + t} d t

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

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

= - ln ∣ 2 - t ∣ + ln ∣ 2 + t ∣

Füge eine Konstante zur Lösung hinzu

= - ln ∣ 2 - t ∣ + ln ∣ 2 + t ∣ + C

\int_{- \infty}^{4} e^{- x} d x

y^{^{''}} + y^{^{'}} = sin^{2} (x)

a re a x = - 2, x = 1, y = 11 x, y = x^{2} - 12

\frac{d}{d x} (400 - 5 x)

\int v (u + v^{2})^{4} d u