integral of 1/(16x^4-8x^2+1)

Lösung

\int \frac{1}{16 x ^{4} - 8 x ^{2} + 1} d x

\frac{1}{8} ln ∣ 2 x + 1 ∣ - \frac{1}{8 ( 2 x + 1 )} - \frac{1}{8} ln ∣ 2 x - 1 ∣ - \frac{1}{8 ( 2 x - 1 )} + C

Schritte zur Lösung

\int \frac{1}{16 x ^{4} - 8 x ^{2} + 1} d x

Ermittle den Partialbruch von

\frac{1}{16 x ^{4} - 8 x ^{2} + 1} : \frac{1}{4 ( 2 x + 1 )} + \frac{1}{4 ( 2 x + 1 ) ^{2}} - \frac{1}{4 ( 2 x - 1 )} + \frac{1}{4 ( 2 x - 1 ) ^{2}}

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

\int \frac{1}{4 ( 2 x + 1 )} d x = \frac{1}{8} ln ∣ 2 x + 1 ∣

\int \frac{1}{4 ( 2 x + 1 ) ^{2}} d x = - \frac{1}{8 ( 2 x + 1 )}

\int \frac{1}{4 ( 2 x - 1 )} d x = \frac{1}{8} ln ∣ 2 x - 1 ∣

\int \frac{1}{4 ( 2 x - 1 ) ^{2}} d x = - \frac{1}{8 ( 2 x - 1 )}

= \frac{1}{8} ln ∣ 2 x + 1 ∣ - \frac{1}{8 ( 2 x + 1 )} - \frac{1}{8} ln ∣ 2 x - 1 ∣ - \frac{1}{8 ( 2 x - 1 )}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{8} ln ∣ 2 x + 1 ∣ - \frac{1}{8 ( 2 x + 1 )} - \frac{1}{8} ln ∣ 2 x - 1 ∣ - \frac{1}{8 ( 2 x - 1 )} + C

n = 0 \sum \infty \frac{4 n}{8 n + 5}

\int_{1}^{2} \frac{ln ( x )}{x ^{3}} d x

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

n or ma l y = x^{2} - 5 x, (5, 0)

\frac{\partial}{\partial z} (y z^{2} - ln (x + z))