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受欢迎的微积分 >

integral of (x+1)/(3(x^2-x+1))

解答

\int \frac{x + 1}{3 ( x ^{2} - x + 1 )} d x

解答

\frac{1}{6} (ln 4 x^{2} - 4 x + 4 + 23 arctan (\frac{2 x - 1}{3})) + C

求解步骤

\int \frac{x + 1}{3 ( x ^{2} - x + 1 )} d x

提出常数:

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

= \frac{1}{3} \cdot \int \frac{x + 1}{x ^{2} - x + 1} d x

对

x^{2} - x + 1

配方:

(x - \frac{1}{2})^{2} + \frac{3}{4}

= \frac{1}{3} \cdot \int \frac{x + 1}{( x - \frac{1}{2} ) ^{2} + \frac{3}{4}} d x

使用换元积分法

= \frac{1}{3} \cdot \int \frac{2 ( 2 u + 3 )}{4 u ^{2} + 3} d u

提出常数:

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

= \frac{1}{3} \cdot 2 \cdot \int \frac{2 u + 3}{4 u ^{2} + 3} d u

乘开

\frac{2 u + 3}{4 u ^{2} + 3} : \frac{2 u}{4 u ^{2} + 3} + \frac{3}{4 u ^{2} + 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

= \frac{1}{3} \cdot 2 (\int \frac{2 u}{4 u ^{2} + 3} d u + \int \frac{3}{4 u ^{2} + 3} d u)

\int \frac{2 u}{4 u ^{2} + 3} d u = \frac{1}{4} ln 4 u^{2} + 3

\int \frac{3}{4 u ^{2} + 3} d u = \frac{3}{2} arctan (\frac{2}{3} u)

= \frac{1}{3} \cdot 2 (\frac{1}{4} ln 4 u^{2} + 3 + \frac{3}{2} arctan (\frac{2}{3} u))

u = x - \frac{1}{2}

代回

= \frac{1}{3} \cdot 2 (\frac{1}{4} ln 4 (x - \frac{1}{2})^{2} + 3 + \frac{3}{2} arctan (\frac{2}{3} (x - \frac{1}{2})))

化简

\frac{1}{3} \cdot 2 (\frac{1}{4} ln 4 (x - \frac{1}{2})^{2} + 3 + \frac{3}{2} arctan (\frac{2}{3} (x - \frac{1}{2}))) : \frac{1}{6} (ln 4 x^{2} - 4 x + 4 + 23 arctan (\frac{2 x - 1}{3}))

= \frac{1}{6} (ln 4 x^{2} - 4 x + 4 + 23 arctan (\frac{2 x - 1}{3}))

解答补常数

= \frac{1}{6} (ln 4 x^{2} - 4 x + 4 + 23 arctan (\frac{2 x - 1}{3})) + C

作图

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