integral of (6x^2+5x+59)/((x+1)(x^2+9))

解答

\int \frac{6 x ^{2} + 5 x + 59}{( x + 1 ) ( x ^{2} + 9 )} d x

6 ln ∣ x + 1 ∣ + \frac{5}{3} arctan (\frac{x}{3}) + C

求解步骤

\int \frac{6 x ^{2} + 5 x + 59}{( x + 1 ) ( x ^{2} + 9 )} d x

将

\frac{6 x ^{2} + 5 x + 59}{( x + 1 ) ( x ^{2} + 9 )}

用部份分式展开:

\frac{6}{x + 1} + \frac{5}{x ^{2} + 9}

= \int \frac{6}{x + 1} + \frac{5}{x ^{2} + 9} 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{6}{x + 1} d x + \int \frac{5}{x ^{2} + 9} d x

\int \frac{6}{x + 1} d x = 6 ln ∣ x + 1 ∣

\int \frac{5}{x ^{2} + 9} d x = \frac{5}{3} arctan (\frac{x}{3})

= 6 ln ∣ x + 1 ∣ + \frac{5}{3} arctan (\frac{x}{3})

解答补常数

= 6 ln ∣ x + 1 ∣ + \frac{5}{3} arctan (\frac{x}{3}) + C

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

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