integral of arctan(sqrt(2x+1))

Lösung

\int arctan (2 x + 1) d x

x arctan (2 x + 1) + arctan (2 x + 1) - \frac{1}{2} 2 x + 1 + 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 U-Substitution an

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

Entferne die Konstante:

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

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

Wende die partielle Integration an

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

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

= \frac{1}{2} \cdot 2 (\frac{1}{2} v^{2} arctan (v) - \frac{1}{2} (- arctan (v) + v))

Ersetze zurück

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

\int csc^{2} (x) x d x

\int x^{3} (x^{2} + 1)^{12} d x

\int (x^{\frac{3}{2}} + 7)^{8} x d x

\int \frac{1}{t ^{2} 361 - t ^{2}} d t

\int y^{2} (7 + y)^{2} d y