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受欢迎的微积分 >

integral of (2x)/(sqrt(x^2+4x+3))

解答

\int \frac{2 x}{x ^{2} + 4 x + 3} d x

解答

2 (x^{2} + 4 x + 3 - 2 ln x^{2} + 4 x + 3 + x + 2) + C

求解步骤

\int \frac{2 x}{x ^{2} + 4 x + 3} d x

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{x}{x ^{2} + 4 x + 3} d x

对

x^{2} + 4 x + 3

配方:

(x + 2)^{2} - 1

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

使用换元积分法

= 2 \cdot \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}

= 2 \cdot \int \frac{u}{u ^{2} - 1} - \frac{2}{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

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

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

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

= 2 (u^{2} - 1 - 2 ln u^{2} - 1 + u)

u = x + 2

代回

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

化简

2 ((x + 2)^{2} - 1 - 2 ln (x + 2)^{2} - 1 + x + 2) : 2 (x^{2} + 4 x + 3 - 2 ln x^{2} + 4 x + 3 + x + 2)

= 2 (x^{2} + 4 x + 3 - 2 ln x^{2} + 4 x + 3 + x + 2)

解答补常数

= 2 (x^{2} + 4 x + 3 - 2 ln x^{2} + 4 x + 3 + x + 2) + C

作图

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