integral of 1/((x+4)^2(x-3))

Lösung

\int \frac{1}{( x + 4 ) ^{2} ( x - 3 )} d x

- \frac{1}{49} ln ∣ x + 4 ∣ + \frac{1}{7 ( x + 4 )} + \frac{1}{49} ln ∣ x - 3 ∣ + C

Schritte zur Lösung

\int \frac{1}{( x + 4 ) ^{2} ( x - 3 )} d x

Ermittle den Partialbruch von

\frac{1}{( x + 4 ) ^{2} ( x - 3 )} : - \frac{1}{49 ( x + 4 )} - \frac{1}{7 ( x + 4 ) ^{2}} + \frac{1}{49 ( x - 3 )}

= \int - \frac{1}{49 ( x + 4 )} - \frac{1}{7 ( x + 4 ) ^{2}} + \frac{1}{49 ( x - 3 )} 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}{49 ( x + 4 )} d x - \int \frac{1}{7 ( x + 4 ) ^{2}} d x + \int \frac{1}{49 ( x - 3 )} d x

\int \frac{1}{49 ( x + 4 )} d x = \frac{1}{49} ln ∣ x + 4 ∣

\int \frac{1}{7 ( x + 4 ) ^{2}} d x = - \frac{1}{7 ( x + 4 )}

\int \frac{1}{49 ( x - 3 )} d x = \frac{1}{49} ln ∣ x - 3 ∣

= - \frac{1}{49} ln ∣ x + 4 ∣ - (- \frac{1}{7 ( x + 4 )}) + \frac{1}{49} ln ∣ x - 3 ∣

Vereinfache

= - \frac{1}{49} ln ∣ x + 4 ∣ + \frac{1}{7 ( x + 4 )} + \frac{1}{49} ln ∣ x - 3 ∣

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{49} ln ∣ x + 4 ∣ + \frac{1}{7 ( x + 4 )} + \frac{1}{49} ln ∣ x - 3 ∣ + C

t an g e n t f (x) = x^{2} + 7, (1, 8)

\frac{d}{d x} (ln (arccos (x)))

2 y^{^{''}} + 5 y^{^{'}} - 18 y = 0

\frac{d x}{d t} = t^{\frac{3}{2}}, x (3) = 7

\frac{d}{d x} (6 x^{- \frac{1}{2}})