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

解答

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

\frac{1}{16} ln ∣ x - 2 ∣ - \frac{1}{16} ln ∣ x + 2 ∣ + \frac{1}{4 ( x + 2 )} + C

求解步骤

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

将

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

用部份分式展开:

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

= \int \frac{1}{16 ( x - 2 )} - \frac{1}{16 ( x + 2 )} - \frac{1}{4 ( x + 2 ) ^{2}} d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{16 ( x - 2 )} d x - \int \frac{1}{16 ( x + 2 )} d x - \int \frac{1}{4 ( x + 2 ) ^{2}} d x

\int \frac{1}{16 ( x - 2 )} d x = \frac{1}{16} ln ∣ x - 2 ∣

\int \frac{1}{16 ( x + 2 )} d x = \frac{1}{16} ln ∣ x + 2 ∣

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

= \frac{1}{16} ln ∣ x - 2 ∣ - \frac{1}{16} ln ∣ x + 2 ∣ - (- \frac{1}{4 ( x + 2 )})

化简

= \frac{1}{16} ln ∣ x - 2 ∣ - \frac{1}{16} ln ∣ x + 2 ∣ + \frac{1}{4 ( x + 2 )}

解答补常数

= \frac{1}{16} ln ∣ x - 2 ∣ - \frac{1}{16} ln ∣ x + 2 ∣ + \frac{1}{4 ( x + 2 )} + C

\int \frac{1}{x ^{2}} 1 - \frac{1}{x} d x

\frac{d y}{d x} = y cos (3 x + 2)

\frac{d y}{d t} + 7 y = e^{5 t}

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

x^{^{''}} + 3 x^{^{'}} + 2 x = 1