integral of (64x^2)/(x^4-40x^2+144)

Lösung

\int \frac{64 x ^{2}}{x ^{4} - 40 x ^{2} + 144} d x

64 (\frac{1}{32} ln ∣ x + 2 ∣ - \frac{1}{32} ln ∣ x - 2 ∣ - \frac{3}{32} ln ∣ x + 6 ∣ + \frac{3}{32} ln ∣ x - 6 ∣) + C

Schritte zur Lösung

\int \frac{64 x ^{2}}{x ^{4} - 40 x ^{2} + 144} d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 64 \cdot \int \frac{x ^{2}}{x ^{4} - 40 x ^{2} + 144} d x

Ermittle den Partialbruch von

\frac{x ^{2}}{x ^{4} - 40 x ^{2} + 144} : \frac{1}{32 ( x + 2 )} - \frac{1}{32 ( x - 2 )} - \frac{3}{32 ( x + 6 )} + \frac{3}{32 ( x - 6 )}

= 64 \cdot \int \frac{1}{32 ( x + 2 )} - \frac{1}{32 ( x - 2 )} - \frac{3}{32 ( x + 6 )} + \frac{3}{32 ( x - 6 )} 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

= 64 (\int \frac{1}{32 ( x + 2 )} d x - \int \frac{1}{32 ( x - 2 )} d x - \int \frac{3}{32 ( x + 6 )} d x + \int \frac{3}{32 ( x - 6 )} d x)

\int \frac{1}{32 ( x + 2 )} d x = \frac{1}{32} ln ∣ x + 2 ∣

\int \frac{1}{32 ( x - 2 )} d x = \frac{1}{32} ln ∣ x - 2 ∣

\int \frac{3}{32 ( x + 6 )} d x = \frac{3}{32} ln ∣ x + 6 ∣

\int \frac{3}{32 ( x - 6 )} d x = \frac{3}{32} ln ∣ x - 6 ∣

= 64 (\frac{1}{32} ln ∣ x + 2 ∣ - \frac{1}{32} ln ∣ x - 2 ∣ - \frac{3}{32} ln ∣ x + 6 ∣ + \frac{3}{32} ln ∣ x - 6 ∣)

Füge eine Konstante zur Lösung hinzu

= 64 (\frac{1}{32} ln ∣ x + 2 ∣ - \frac{1}{32} ln ∣ x - 2 ∣ - \frac{3}{32} ln ∣ x + 6 ∣ + \frac{3}{32} ln ∣ x - 6 ∣) + C

\int 3 sec^{3} (x) d x

d er i v a t i v e - 7 x^{\frac{2}{3}} - 2 + 3 x^{\frac{7}{4}}

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

d er i v a t i v e f (x) = 2 x^{2} - 196 ln (x)

\int \frac{1}{( 16 - t ^{2} ) ^{\frac{3}{2}}} d t