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

Lösung

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

ln ∣ x ∣ - \frac{1}{2} ln 4 x^{2} + 1 + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ - \frac{1}{2} ln 4 x^{2} + 1

Füge eine Konstante zur Lösung hinzu

= ln ∣ x ∣ - \frac{1}{2} ln 4 x^{2} + 1 + C

s im pl i f y \frac{- a}{x ^{2}} + \frac{a}{( x - 2 ) ^{2}}

\int_{0}^{2} \frac{1}{4 - 2 x} d x

\int_{1}^{5} (2 + 3 x - x^{2}) d x

t an g e n t \frac{1}{10 x}

\int 2 e^{- t} d t