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受欢迎的微积分 >

integral of (w-1)/(2w^2-2w+1)

解答

\int \frac{w - 1}{2 w ^{2} - 2 w + 1} d w

解答

\frac{1}{4} ln 4 w^{2} - 4 w + 2 - \frac{1}{2} arctan (2 w - 1) + C

求解步骤

\int \frac{w - 1}{2 w ^{2} - 2 w + 1} d w

对

2 w^{2} - 2 w + 1

配方:

2 (w - \frac{1}{2})^{2} + \frac{1}{2}

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

使用换元积分法

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

乘开

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

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

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

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

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

代回

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

化简

\frac{1}{4} ln 4 (w - \frac{1}{2})^{2} + 1 - \frac{1}{2} arctan (2 (w - \frac{1}{2})) : \frac{1}{4} ln 4 w^{2} - 4 w + 2 - \frac{1}{2} arctan (2 w - 1)

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

解答补常数

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

作图

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