integral of 5/(x(x+4)^2)

解答

\int \frac{5}{x ( x + 4 ) ^{2}} d x

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

求解步骤

\int \frac{5}{x ( x + 4 ) ^{2}} d x

提出常数:

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

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

将

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

用部份分式展开:

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

= 5 \cdot \int \frac{1}{16 x} - \frac{1}{16 ( x + 4 )} - \frac{1}{4 ( x + 4 ) ^{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

= 5 (\int \frac{1}{16 x} d x - \int \frac{1}{16 ( x + 4 )} d x - \int \frac{1}{4 ( x + 4 ) ^{2}} d x)

\int \frac{1}{16 x} d x = \frac{1}{16} ln ∣ x ∣

\int \frac{1}{16 ( x + 4 )} d x = \frac{1}{16} ln ∣ x + 4 ∣

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

= 5 (\frac{1}{16} ln ∣ x ∣ - \frac{1}{16} ln ∣ x + 4 ∣ - (- \frac{1}{4 ( x + 4 )}))

化简

= 5 (\frac{1}{16} ln ∣ x ∣ - \frac{1}{16} ln ∣ x + 4 ∣ + \frac{1}{4 ( x + 4 )})

解答补常数

= 5 (\frac{1}{16} ln ∣ x ∣ - \frac{1}{16} ln ∣ x + 4 ∣ + \frac{1}{4 ( x + 4 )}) + C

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