integral of (x^2+3)/(x+1)

解答

\int \frac{x ^{2} + 3}{x + 1} d x

\frac{( x + 1 ) ^{2}}{2} - 2 (x + 1) + 4 ln ∣ x + 1 ∣ + C

求解步骤

\int \frac{x ^{2} + 3}{x + 1} d x

使用换元积分法

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

乘开

\frac{u ^{2} - 2 u + 4}{u} : u - 2 + \frac{4}{u}

= \int u - 2 + \frac{4}{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 u d u - \int 2 d u + \int \frac{4}{u} d u

\int u d u = \frac{u ^{2}}{2}

\int 2 d u = 2 u

\int \frac{4}{u} d u = 4 ln ∣ u ∣

= \frac{u ^{2}}{2} - 2 u + 4 ln ∣ u ∣

u = x + 1

代回

= \frac{( x + 1 ) ^{2}}{2} - 2 (x + 1) + 4 ln ∣ x + 1 ∣

解答补常数

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

\frac{d}{d x} (\frac{x ^{4}}{4} + \frac{1}{8 x ^{2}})

\int_{0}^{2} x^{2} ln (x) d x

\int_{2}^{5} x^{6} d x

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

\int e^{c o s (7 t)} sin (7 t) d t