integral of (e^{arctan(x)}+1)/(1+x^2)

Lösung

\int \frac{e ^{a r c t a n (x)} + 1}{1 + x ^{2}} d x

e^{a r c t a n (x)} + arctan (x) + C

Schritte zur Lösung

\int \frac{e ^{a r c t a n (x)} + 1}{1 + x ^{2}} d x

Multipliziere aus

\frac{e ^{a r c t a n (x)} + 1}{1 + x ^{2}} : \frac{e ^{a r c t a n (x)}}{1 + x ^{2}} + \frac{1}{1 + x ^{2}}

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{e ^{a r c t a n (x)}}{1 + x ^{2}} d x + \int \frac{1}{1 + x ^{2}} d x

\int \frac{e ^{a r c t a n (x)}}{1 + x ^{2}} d x = e^{a r c t a n (x)}

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

= e^{a r c t a n (x)} + arctan (x)

Füge eine Konstante zur Lösung hinzu

= e^{a r c t a n (x)} + arctan (x) + C

x \to - 1 - lim (\frac{x}{∣ x - 1 ∣})

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