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受欢迎的微积分 >

integral of ((2z-1))/(2z^2-6z+5)

解答

\int \frac{( 2 z - 1 )}{2 z ^{2} - 6 z + 5} d z

解答

\frac{1}{2} ln 4 z^{2} - 12 z + 10 + 2 arctan (2 z - 3) + C

求解步骤

\int \frac{2 z - 1}{2 z ^{2} - 6 z + 5} d z

使用换元积分法

= \int \frac{u}{u ^{2} - 4 u + 5} d u

对

u^{2} - 4 u + 5

配方:

(u - 2)^{2} + 1

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

使用换元积分法

= \int \frac{v + 2}{v ^{2} + 1} d v

乘开

\frac{v + 2}{v ^{2} + 1} : \frac{v}{v ^{2} + 1} + \frac{2}{v ^{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{v}{v ^{2} + 1} d v + \int \frac{2}{v ^{2} + 1} d v

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

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

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

代回

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

化简

\frac{1}{2} ln (2 z - 1 - 2)^{2} + 1 + 2 arctan (2 z - 1 - 2) : \frac{1}{2} ln 4 z^{2} - 12 z + 10 + 2 arctan (2 z - 3)

= \frac{1}{2} ln 4 z^{2} - 12 z + 10 + 2 arctan (2 z - 3)

解答补常数

= \frac{1}{2} ln 4 z^{2} - 12 z + 10 + 2 arctan (2 z - 3) + C

作图

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