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

解答

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

\frac{1}{25} ln ∣ x + 4 ∣ - \frac{1}{25} ln ∣ x - 1 ∣ - \frac{1}{5 ( x - 1 )} + C

求解步骤

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

将

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

用部份分式展开:

\frac{1}{25 ( x + 4 )} - \frac{1}{25 ( x - 1 )} + \frac{1}{5 ( x - 1 ) ^{2}}

= \int \frac{1}{25 ( x + 4 )} - \frac{1}{25 ( x - 1 )} + \frac{1}{5 ( x - 1 ) ^{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}{25 ( x + 4 )} d x - \int \frac{1}{25 ( x - 1 )} d x + \int \frac{1}{5 ( x - 1 ) ^{2}} d x

\int \frac{1}{25 ( x + 4 )} d x = \frac{1}{25} ln ∣ x + 4 ∣

\int \frac{1}{25 ( x - 1 )} d x = \frac{1}{25} ln ∣ x - 1 ∣

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

= \frac{1}{25} ln ∣ x + 4 ∣ - \frac{1}{25} ln ∣ x - 1 ∣ - \frac{1}{5 ( x - 1 )}

解答补常数

= \frac{1}{25} ln ∣ x + 4 ∣ - \frac{1}{25} ln ∣ x - 1 ∣ - \frac{1}{5 ( x - 1 )} + C

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