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受欢迎的微积分 >

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

解答

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

解答

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

求解步骤

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

对

4 x^{2} + 4 x + 5

配方:

4 (x + \frac{1}{2})^{2} + 4

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

使用换元积分法

= \int \frac{2 u - 1}{4 u ^{2} + 1} d u

提出常数:

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

= \frac{1}{4} \cdot \int \frac{2 u - 1}{u ^{2} + 1} d u

乘开

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

= \frac{1}{4} \cdot \int \frac{2 u}{u ^{2} + 1} - \frac{1}{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

= \frac{1}{4} (\int \frac{2 u}{u ^{2} + 1} d u - \int \frac{1}{u ^{2} + 1} d u)

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

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

= \frac{1}{4} (2 u^{2} + 1 - ln u^{2} + 1 + u)

u = x + \frac{1}{2}

代回

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

化简

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

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

解答补常数

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

作图

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