integral of (x+1)/((x^2+4x+5)^2)

Lösung

\int \frac{x + 1}{( x ^{2} + 4 x + 5 ) ^{2}} d x

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

Schritte zur Lösung

\int \frac{x + 1}{( x ^{2} + 4 x + 5 ) ^{2}} d x

Wende die partielle Integration an

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

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

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

d er i v a t i v e f (x) = (4 a + 1)^{2}

s l o p e y = \frac{x ln ( x )}{6} + \frac{2 x}{3}

\int 5 x e^{2 x} d x

d er i v a t i v e arcsec (4 x)

d er i v a t i v e f (x) = 2 x + 7