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integral of (16x^2)/((x-12)(x+4)^2)

解答

\int \frac{16 x ^{2}}{( x - 12 ) ( x + 4 ) ^{2}} d x

解答

9 ln ∣ x - 12 ∣ + 7 ln ∣ x + 4 ∣ + \frac{16}{x + 4} + C

求解步骤

\int \frac{16 x ^{2}}{( x - 12 ) ( x + 4 ) ^{2}} d x

提出常数:

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

= 16 \cdot \int \frac{x ^{2}}{( x - 12 ) ( x + 4 ) ^{2}} d x

将

\frac{x ^{2}}{( x - 12 ) ( x + 4 ) ^{2}}

用部份分式展开:

\frac{9}{16 ( x - 12 )} + \frac{7}{16 ( x + 4 )} - \frac{1}{( x + 4 ) ^{2}}

= 16 \cdot \int \frac{9}{16 ( x - 12 )} + \frac{7}{16 ( x + 4 )} - \frac{1}{( 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

= 16 (\int \frac{9}{16 ( x - 12 )} d x + \int \frac{7}{16 ( x + 4 )} d x - \int \frac{1}{( x + 4 ) ^{2}} d x)

\int \frac{9}{16 ( x - 12 )} d x = \frac{9}{16} ln ∣ x - 12 ∣

\int \frac{7}{16 ( x + 4 )} d x = \frac{7}{16} ln ∣ x + 4 ∣

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

= 16 (\frac{9}{16} ln ∣ x - 12 ∣ + \frac{7}{16} ln ∣ x + 4 ∣ - (- \frac{1}{x + 4}))

化简

16 (\frac{9}{16} ln ∣ x - 12 ∣ + \frac{7}{16} ln ∣ x + 4 ∣ - (- \frac{1}{x + 4})) : 9 ln ∣ x - 12 ∣ + 7 ln ∣ x + 4 ∣ + \frac{16}{x + 4}

= 9 ln ∣ x - 12 ∣ + 7 ln ∣ x + 4 ∣ + \frac{16}{x + 4}

解答补常数

= 9 ln ∣ x - 12 ∣ + 7 ln ∣ x + 4 ∣ + \frac{16}{x + 4} + C

作图

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