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受欢迎的微积分 >

integral of (x-1)/(x^2-4x+5)

解答

\int \frac{x - 1}{x ^{2} - 4 x + 5} d x

解答

\frac{1}{2} ln x^{2} - 4 x + 5 + arctan (x - 2) + C

求解步骤

\int \frac{x - 1}{x ^{2} - 4 x + 5} d x

对

x^{2} - 4 x + 5

配方:

(x - 2)^{2} + 1

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

使用换元积分法

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

乘开

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

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

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

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

u = x - 2

代回

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

化简

\frac{1}{2} ln (x - 2)^{2} + 1 + arctan (x - 2) : \frac{1}{2} ln x^{2} - 4 x + 5 + arctan (x - 2)

= \frac{1}{2} ln x^{2} - 4 x + 5 + arctan (x - 2)

解答补常数

= \frac{1}{2} ln x^{2} - 4 x + 5 + arctan (x - 2) + C

作图

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