integral of arctan(2x+1)

Lösung

\int arctan (2 x + 1) d x

\frac{1}{4} (2 arctan (2 x + 1) (2 x + 1) - ln 4 x^{2} + 4 x + 2) + C

Schritte zur Lösung

\int arctan (2 x + 1) d x

Wende U-Substitution an

= \int arctan (u) \frac{1}{2} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int arctan (u) d u

Wende die partielle Integration an

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

\int \frac{u}{u ^{2} + 1} d u = \frac{1}{2} ln u^{2} + 1

= \frac{1}{2} (u arctan (u) - \frac{1}{2} ln u^{2} + 1)

Setze in

u = 2 x + 1

ein

= \frac{1}{2} ((2 x + 1) arctan (2 x + 1) - \frac{1}{2} ln (2 x + 1)^{2} + 1)

Vereinfache

\frac{1}{2} ((2 x + 1) arctan (2 x + 1) - \frac{1}{2} ln (2 x + 1)^{2} + 1) : \frac{1}{4} (2 arctan (2 x + 1) (2 x + 1) - ln 4 x^{2} + 4 x + 2)

= \frac{1}{4} (2 arctan (2 x + 1) (2 x + 1) - ln 4 x^{2} + 4 x + 2)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4} (2 arctan (2 x + 1) (2 x + 1) - ln 4 x^{2} + 4 x + 2) + C

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