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受欢迎的微积分 >

integral of (10y+11)/(4y^2-4y+5)

解答

\int \frac{10 y + 11}{4 y ^{2} - 4 y + 5} d y

解答

\frac{1}{4} (5 ln y^{2} - y + \frac{5}{4} + 16 arctan (y - \frac{1}{2})) + C

求解步骤

\int \frac{10 y + 11}{4 y ^{2} - 4 y + 5} d y

对

4 y^{2} - 4 y + 5

配方:

4 (y - \frac{1}{2})^{2} + 4

= \int \frac{10 y + 11}{4 ( y - \frac{1}{2} ) ^{2} + 4} d y

使用换元积分法

= \int \frac{5 u + 8}{2 ( 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}{2} \cdot \int \frac{5 u + 8}{u ^{2} + 1} d u

乘开

\frac{5 u + 8}{u ^{2} + 1} : \frac{5 u}{u ^{2} + 1} + \frac{8}{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

= \frac{1}{2} (\int \frac{5 u}{u ^{2} + 1} d u + \int \frac{8}{u ^{2} + 1} d u)

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

\int \frac{8}{u ^{2} + 1} d u = 8 arctan (u)

= \frac{1}{2} (\frac{5}{2} ln u^{2} + 1 + 8 arctan (u))

u = y - \frac{1}{2}

代回

= \frac{1}{2} (\frac{5}{2} ln (y - \frac{1}{2})^{2} + 1 + 8 arctan (y - \frac{1}{2}))

化简

\frac{1}{2} (\frac{5}{2} ln (y - \frac{1}{2})^{2} + 1 + 8 arctan (y - \frac{1}{2})) : \frac{1}{4} (5 ln y^{2} - y + \frac{5}{4} + 16 arctan (y - \frac{1}{2}))

= \frac{1}{4} (5 ln y^{2} - y + \frac{5}{4} + 16 arctan (y - \frac{1}{2}))

解答补常数

= \frac{1}{4} (5 ln y^{2} - y + \frac{5}{4} + 16 arctan (y - \frac{1}{2})) + C

作图

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