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受欢迎的微积分 >

integral of (x-1)/(4x^2-4x+3)

解答

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

解答

\frac{1}{8} (ln 4 x^{2} - 4 x + 3 - 2 arctan (2 (x - \frac{1}{2}))) + C

求解步骤

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

对

4 x^{2} - 4 x + 3

配方:

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

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

使用换元积分法

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

提出常数:

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

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

乘开

\frac{2 u - 1}{4 u ^{2} + 2} : \frac{2 u}{4 u ^{2} + 2} - \frac{1}{4 u ^{2} + 2}

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

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

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

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

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

代回

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

化简

\frac{1}{2} (\frac{1}{4} ln 4 (x - \frac{1}{2})^{2} + 2 - \frac{1}{2 2} arctan (2 (x - \frac{1}{2}))) : \frac{1}{8} (ln 4 x^{2} - 4 x + 3 - 2 arctan (2 (x - \frac{1}{2})))

= \frac{1}{8} (ln 4 x^{2} - 4 x + 3 - 2 arctan (2 (x - \frac{1}{2})))

解答补常数

= \frac{1}{8} (ln 4 x^{2} - 4 x + 3 - 2 arctan (2 (x - \frac{1}{2}))) + C

作图

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