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

解答

\int \frac{( 5 t - 2 )}{t + 1} d t

5 (t + 1) - 7 ln ∣ t + 1 ∣ + C

求解步骤

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

使用换元积分法

= \int \frac{5 u - 7}{u} d u

乘开

\frac{5 u - 7}{u} : 5 - \frac{7}{u}

= \int 5 - \frac{7}{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 5 d u - \int \frac{7}{u} d u

\int 5 d u = 5 u

\int \frac{7}{u} d u = 7 ln ∣ u ∣

= 5 u - 7 ln ∣ u ∣

u = t + 1

代回

= 5 (t + 1) - 7 ln ∣ t + 1 ∣

解答补常数

= 5 (t + 1) - 7 ln ∣ t + 1 ∣ + C

\frac{d y}{d x} = \frac{( x ^{2} - 1 )}{y ^{2}}

\int \frac{e ^{x}}{3} - 4 x^{3} + \frac{6}{x} d x

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

n = 1 \sum \infty \frac{3}{n ^{2}}

(x y^{2} + x) d x + (y - x^{2} y) d y = 0