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

Lösung

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

\frac{1}{4 5} (ln \frac{2}{5} x + 1 - ln \frac{2}{5} x - 1) - \frac{1}{8} ln 5 - 4 x^{2} + C

Schritte zur Lösung

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

Multipliziere aus

\frac{1 + x}{5 - 4 x ^{2}} : \frac{1}{5 - 4 x ^{2}} + \frac{x}{5 - 4 x ^{2}}

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}{5 - 4 x ^{2}} d x + \int \frac{x}{5 - 4 x ^{2}} d x

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

\int \frac{x}{5 - 4 x ^{2}} d x = - \frac{1}{8} ln 5 - 4 x^{2}

= \frac{1}{4 5} (ln \frac{2}{5} x + 1 - ln \frac{2}{5} x - 1) - \frac{1}{8} ln 5 - 4 x^{2}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4 5} (ln \frac{2}{5} x + 1 - ln \frac{2}{5} x - 1) - \frac{1}{8} ln 5 - 4 x^{2} + C

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