integral of arctan(x^2)

Lösung

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

x arctan (x^{2}) - 2 (- \frac{1}{4 2} (ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1)) + \frac{1}{4 2} (ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1))) + C

Schritte zur Lösung

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

Wende die partielle Integration an

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

\int \frac{2 x ^{2}}{x ^{4} + 1} d x = 2 (- \frac{1}{4 2} (ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1)) + \frac{1}{4 2} (ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1)))

= x arctan (x^{2}) - 2 (- \frac{1}{4 2} (ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1)) + \frac{1}{4 2} (ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1)))

Füge eine Konstante zur Lösung hinzu

= x arctan (x^{2}) - 2 (- \frac{1}{4 2} (ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1)) + \frac{1}{4 2} (ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1))) + C

x \to - 5 + lim (\frac{- 3}{x ^{2} - 25})

\int (3 x - 5) d x

\int x^{2} \cdot e^{- x^{2}} d x

a re a 2 - x^{2}, x

d er i v a t i v e - 4.8 x^{\frac{1}{4}}