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受欢迎的微积分 >

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

解答

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

解答

2 x^{2} + 4 x + 5 - 5 ln x^{2} + 4 x + 5 + x + 2 + C

求解步骤

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

对

x^{2} + 4 x + 5

配方:

(x + 2)^{2} + 1

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

使用换元积分法

= \int \frac{2 u - 5}{u ^{2} + 1} d u

乘开

\frac{2 u - 5}{u ^{2} + 1} : \frac{2 u}{u ^{2} + 1} - \frac{5}{u ^{2} + 1}

= \int \frac{2 u}{u ^{2} + 1} - \frac{5}{u ^{2} + 1} d u

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

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

\int \frac{5}{u ^{2} + 1} d u = 5 ln u^{2} + 1 + u

= 2 u^{2} + 1 - 5 ln u^{2} + 1 + u

u = x + 2

代回

= 2 (x + 2)^{2} + 1 - 5 ln (x + 2)^{2} + 1 + x + 2

化简

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

= 2 x^{2} + 4 x + 5 - 5 ln x^{2} + 4 x + 5 + x + 2

解答补常数

= 2 x^{2} + 4 x + 5 - 5 ln x^{2} + 4 x + 5 + x + 2 + C

作图

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