integral of ((4x+3))/(x^2+1)

Lösung

\int \frac{( 4 x + 3 )}{x ^{2} + 1} d x

2 ln x^{2} + 1 + 3 arctan (x) + C

Schritte zur Lösung

\int \frac{4 x + 3}{x ^{2} + 1} d x

Multipliziere aus

\frac{4 x + 3}{x ^{2} + 1} : \frac{4 x}{x ^{2} + 1} + \frac{3}{x ^{2} + 1}

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{4 x}{x ^{2} + 1} d x + \int \frac{3}{x ^{2} + 1} d x

\int \frac{4 x}{x ^{2} + 1} d x = 2 ln x^{2} + 1

\int \frac{3}{x ^{2} + 1} d x = 3 arctan (x)

= 2 ln x^{2} + 1 + 3 arctan (x)

Füge eine Konstante zur Lösung hinzu

= 2 ln x^{2} + 1 + 3 arctan (x) + C

x \to \infty lim (5 + \frac{4}{x})

d er i v a t i v e y = (5 t - 1) (2 t - 3)^{- 1}

n = 1 \sum \infty \frac{2}{n ( n + 1 )}

a re a y = 7 x, y = 44 - x^{2}

n = 2 \sum \infty \frac{1}{ln ( n )}