integral of 1-1/(x+1)

解答

\int 1 - \frac{1}{x + 1} d x

x - ln ∣ x + 1 ∣ + C

求解步骤

\int 1 - \frac{1}{x + 1} d x

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 1 d x - \int \frac{1}{x + 1} d x

\int 1 d x = x

\int \frac{1}{x + 1} d x = ln ∣ x + 1 ∣

= x - ln ∣ x + 1 ∣

解答补常数

= x - ln ∣ x + 1 ∣ + C

\int \frac{3 x}{x ^{2} - 4} d x

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

y^{^{'}} = (x - 5) e^{- 2 y}

\int_{1}^{2} \frac{v ^{4} + 5 v ^{7}}{v ^{3}} d v

l a pl a ce t r an s f or m - t^{2} δ (t - 2)