integral of (t^3-8t+1)/(8t^4)

解答

\int \frac{t ^{3} - 8 t + 1}{8 t ^{4}} d t

\frac{1}{8} (ln ∣ t ∣ + \frac{4}{t ^{2}} - \frac{1}{3 t ^{3}}) + C

求解步骤

\int \frac{t ^{3} - 8 t + 1}{8 t ^{4}} d t

提出常数:

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

= \frac{1}{8} \cdot \int \frac{t ^{3} - 8 t + 1}{t ^{4}} d t

乘开

\frac{t ^{3} - 8 t + 1}{t ^{4}} : \frac{1}{t} - \frac{8}{t ^{3}} + \frac{1}{t ^{4}}

= \frac{1}{8} \cdot \int \frac{1}{t} - \frac{8}{t ^{3}} + \frac{1}{t ^{4}} 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

= \frac{1}{8} (\int \frac{1}{t} d t - \int \frac{8}{t ^{3}} d t + \int \frac{1}{t ^{4}} d t)

\int \frac{1}{t} d t = ln ∣ t ∣

\int \frac{8}{t ^{3}} d t = - \frac{4}{t ^{2}}

\int \frac{1}{t ^{4}} d t = - \frac{1}{3 t ^{3}}

= \frac{1}{8} (ln ∣ t ∣ - (- \frac{4}{t ^{2}}) - \frac{1}{3 t ^{3}})

化简

= \frac{1}{8} (ln ∣ t ∣ + \frac{4}{t ^{2}} - \frac{1}{3 t ^{3}})

解答补常数

= \frac{1}{8} (ln ∣ t ∣ + \frac{4}{t ^{2}} - \frac{1}{3 t ^{3}}) + C

\int (\frac{x}{2} - \frac{1}{x} + \frac{3}{x ^{3}} - \frac{4}{x ^{4}}) d x

\int cos^{18} (θ) sin (θ) d θ

\int 3 r^{2} d r

\int 4 x e^{x + 2} d x

\int (5 x^{2}) e^{- x^{3}} d x