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

解答

\int (\frac{4 - t + t ^{2}}{t ^{2}}) d t

- \frac{4}{t} - ln ∣ t ∣ + t + C

求解步骤

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

乘开

\frac{4 - t + t ^{2}}{t ^{2}} : \frac{4}{t ^{2}} - \frac{1}{t} + 1

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

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

\int \frac{4}{t ^{2}} d t = - \frac{4}{t}

\int \frac{1}{t} d t = ln ∣ t ∣

\int 1 d t = t

= - \frac{4}{t} - ln ∣ t ∣ + t

解答补常数

= - \frac{4}{t} - ln ∣ t ∣ + t + C

\frac{d}{d x} (ln (x^{2} + 20))

\frac{\partial}{\partial x} (\frac{2 y}{x ^{2} + 1})

x \to 3 lim (\frac{x - 2}{x + 2})

x \to 1800 lim (28 π^{2})

d er i v a t i v e \frac{x ^{2} - 8 x + 5}{x ^{2} + 8}