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受欢迎的微积分 >

integral of (2x)/(x^2+4x+6)

解答

\int \frac{2 x}{x ^{2} + 4 x + 6} d x

解答

ln \frac{1}{2} (x + 2)^{2} + 1 - 22 arctan (\frac{1}{2} (x + 2)) + C

求解步骤

\int \frac{2 x}{x ^{2} + 4 x + 6} 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 + 6} d x

对

x^{2} + 4 x + 6

配方:

(x + 2)^{2} + 2

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

使用换元积分法

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

使用换元积分法

= 2 \cdot \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

= 2 (\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)

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

代回

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

化简

2 (\frac{1}{2} ln (\frac{x + 2}{2})^{2} + 1 - 2 arctan (\frac{x + 2}{2})) : ln \frac{1}{2} (x + 2)^{2} + 1 - 22 arctan (\frac{1}{2} (x + 2))

= ln \frac{1}{2} (x + 2)^{2} + 1 - 22 arctan (\frac{1}{2} (x + 2))

解答补常数

= ln \frac{1}{2} (x + 2)^{2} + 1 - 22 arctan (\frac{1}{2} (x + 2)) + C

作图

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