integral of (7t-5)/(t+1)

解答

\int \frac{7 t - 5}{t + 1} d t

7 (t + 1) - 12 ln ∣ t + 1 ∣ + C

求解步骤

\int \frac{7 t - 5}{t + 1} d t

使用换元积分法

= \int \frac{7 u - 12}{u} d u

乘开

\frac{7 u - 12}{u} : 7 - \frac{12}{u}

= \int 7 - \frac{12}{u} 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

= \int 7 d u - \int \frac{12}{u} d u

\int 7 d u = 7 u

\int \frac{12}{u} d u = 12 ln ∣ u ∣

= 7 u - 12 ln ∣ u ∣

u = t + 1

代回

= 7 (t + 1) - 12 ln ∣ t + 1 ∣

解答补常数

= 7 (t + 1) - 12 ln ∣ t + 1 ∣ + C

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\frac{\partial}{\partial x} (y^{2} - y)

d er i v a t i v e f (r) = π r^{2}

\frac{d}{d x} (e^{c x})