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

解答

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

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

求解步骤

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

使用换元积分法

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

乘开

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

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

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

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

u = 2 t

代回

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

(2 t)^{2} = 4 t^{2}

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

解答补常数

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

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