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

Lösung

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

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

Schritte zur Lösung

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

Multipliziere aus

\frac{x + 2}{x ^{2} + 1} : \frac{x}{x ^{2} + 1} + \frac{2}{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{x}{x ^{2} + 1} d x + \int \frac{2}{x ^{2} + 1} d x

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

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

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

Füge eine Konstante zur Lösung hinzu

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

\int \frac{1}{x 64 x ^{2} - 1} d x

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

\int \frac{3}{( x - 6 ) ( x - 3 )} d x

\int (m sec (θ) - tan (θ))^{2} d θ

\int \frac{3 x - 2}{5 x ^{2} - 3 x + 1} d x