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

Lösung

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

ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1 + C

Schritte zur Lösung

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

Vereinfache

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

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

Ermittle den Partialbruch von

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

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1

Füge eine Konstante zur Lösung hinzu

= ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1 + C

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

\int x^{3} e x^{4} d x

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

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

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