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

解答

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

\frac{1}{2} ln \frac{x ^{2}}{9} + 1 + 3 arctan (\frac{x}{3}) + C

求解步骤

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

使用换元积分法

= \int \frac{u + 3}{u ^{2} + 1} d u

乘开

\frac{u + 3}{u ^{2} + 1} : \frac{u}{u ^{2} + 1} + \frac{3}{u ^{2} + 1}

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{u}{u ^{2} + 1} d u + \int \frac{3}{u ^{2} + 1} d u

\int \frac{u}{u ^{2} + 1} d u = \frac{1}{2} ln u^{2} + 1

\int \frac{3}{u ^{2} + 1} d u = 3 arctan (u)

= \frac{1}{2} ln u^{2} + 1 + 3 arctan (u)

u = \frac{x}{3}

代回

= \frac{1}{2} ln (\frac{x}{3})^{2} + 1 + 3 arctan (\frac{x}{3})

(\frac{x}{3})^{2} = \frac{x ^{2}}{9}

= \frac{1}{2} ln \frac{x ^{2}}{9} + 1 + 3 arctan (\frac{x}{3})

解答补常数

= \frac{1}{2} ln \frac{x ^{2}}{9} + 1 + 3 arctan (\frac{x}{3}) + C

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