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受欢迎的微积分 >

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

解答

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

解答

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

求解步骤

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

对

x^{2} - 2 x + 2

配方:

(x - 1)^{2} + 1

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

使用换元积分法

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

乘开

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

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

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

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

u = x - 1

代回

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

化简

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

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

解答补常数

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

作图

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