integral of (x^2-9)/(x+3)

解答

\int \frac{x ^{2} - 9}{x + 3} d x

\frac{1}{2} (x + 3)^{2} - 6 x - 18 + C

求解步骤

\int \frac{x ^{2} - 9}{x + 3} d x

乘开

\frac{x ^{2} - 9}{x + 3} : \frac{x ^{2}}{x + 3} - \frac{9}{x + 3}

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{x ^{2}}{x + 3} d x - \int \frac{9}{x + 3} d x

\int \frac{x ^{2}}{x + 3} d x = \frac{( x + 3 ) ^{2}}{2} - 6 (x + 3) + 9 ln ∣ x + 3 ∣

\int \frac{9}{x + 3} d x = 9 ln ∣ x + 3 ∣

= \frac{( x + 3 ) ^{2}}{2} - 6 (x + 3) + 9 ln ∣ x + 3 ∣ - 9 ln ∣ x + 3 ∣

化简

\frac{( x + 3 ) ^{2}}{2} - 6 (x + 3) + 9 ln ∣ x + 3 ∣ - 9 ln ∣ x + 3 ∣ : \frac{1}{2} (x + 3)^{2} - 6 x - 18

= \frac{1}{2} (x + 3)^{2} - 6 x - 18

解答补常数

= \frac{1}{2} (x + 3)^{2} - 6 x - 18 + C

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n = 0 \sum \infty n e^{n}

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